I was wondering about this. I know it is possible to visualize the quotient group $\mathbb{R}/\mathbb{Z}$ as a circle, and if you consider these as "topological groups", then this group (not topological) quotient is topologically equivalent to a circle.
But then, what does $\mathbb{R}/\mathbb{Q}$ look like?
So, you say that the group (not topological) quotient of $\mathbb{R}/\mathbb{Z}$ is topologically equivalent (i.e., homeomorphic) to the circle. However, this doesn't make any sense unless you have a topology on $\mathbb{R}/\mathbb{Z}$! More the point is that a topological group like $\mathbb{R}$ has both a topological structure and a group structure. Now, when you form the group quotient $\mathbb{R}/\mathbb{Z}$, it can be given a topological space in a natural way, in particular, via the quotient topology. Notice that when we do this we again get a topological group (i.e., the quotient group operations are continuous with respect to the quotient topology). Furthermore, the quotient $\mathbb{R}/\mathbb{Z}$ (as a topological space) is homeomorphic to the circle.
Now, in the case of your question, the quotient topology on $\mathbb{R}/\mathbb{Q}$ is the trivial topology. This is not hard to prove since preimages of open sets must be open and saturated. Thus if such a preimage is nonempty, it contains an open interval, and since it is saturated, it must contain all real numbers which differ by a rational from a point in this interval. It is then easy to see that this set must be all of $\mathbb{R}$. Thus the only saturated open sets of $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$ itself. Hence the quotient topology is trivial. Furthermore, it is trivial that any map into a space with the trivial topology is continuous, so the quotient group operations on $\mathbb{R}/\mathbb{Q}$ are again continuous. So we again have a topological group, albeit not a very interesting one because it isn't very interesting as a topological space. As far as what this space "looks" like, it is similar to a one point space for the reason Ricky mentioned in the comments. However, it is not really easy to visualize since it is not homeomorphic to any subspace of $\mathbb{R}^n$ equipped with the subspace topology (because it is not Hausdorff, or any one of a number of other reasons).
Edit: I should have added that whenever you have a topological group and form the quotient in the way we did above the result is always a topological group. However, unless the original normal subgroup is closed, the resulting quotient group will not even be $T_0$ as a topological space. Thus it is only really interesting to form the quotient when the set by which you quotient out is closed. This explains why $\mathbb{R}/\mathbb{Z}$ is interesting as a topological group, but $\mathbb{R}/\mathbb{Q}$ is not.
If you ignore topology, it's pretty much the same as $\mathbf R$.
Notice that $\mathbf R$ is a $\mathfrak c$-dimensional vector space over $\mathbf Q$, of which $\bf Q$ is a one-dimensional subspace. Taking the quotient $\bf R/\bf Q$ is actually taking the quotient of a $\mathfrak c$-dimensional vector space by a one-dimensional subspace, which is again a vector space, and is still $\mathfrak c$-dimensional (because $1<\mathfrak c$ ;) ), so it is isomorphic to $\bf R$ as a vector space over $\bf Q$, and in particular as a group.
It really depends on what you think about as visualizing.
The group $\mathbb Z$ is discrete, so between two successive points there is a part which looks a bit like $\mathbb R$. The result, if so, is somewhat close to being $\mathbb R$.
On the other hand, $\mathbb Q$ is a dense subgroup of $\mathbb R$. This means that it gets a lot messier. Not without a good reason too, we can usually imagine things which have shape, things which can be measured.
Any set of representatives for $\mathbb R/\mathbb Q$ cannot be measured. This tells you that it is practically impossible to visualize this quotient in the same sense that we would imagine a circle, a ball, or even if we try really hard and we imagine a four-dimensional space.
Furthermore, using the axiom of choice we can create such set of representatives; however without the axiom of choice this quotient might not even be linearly ordered. Namely, it forms a set which cannot be linearly ordered. In contrast, $\mathbb R/\mathbb Z$ is a circle, or a half-open interval (where we identify the endpoints), even without the axiom of choice.
This tells you even more: you need the axiom of choice to impose an order on this set. Just a linear order, not even a well-order. Therefore imagining this as a linearly ordered set is even harder than we may believe at first.
My suggestion is not to try and visualize it. Accept this as a formal object which you can understand to some extent, but not see. Move on with this. Eventually, after running into infinitary objects ($\ell^2$, for example) and succeeding in visualizing those -- come back to this one, then you might be able to pull this off.
The quotient group R/Q is similar to R/Z in some respects, but is quite different and, I think, impossible to visualize in the way R/Z is. First note that if p is a rational number, then it's equivalence class (i.e. coset generated by p) in R/Q, denoted [p] is equal to [0]. That is, all rationals collapse to the single coset Q. Now, note that if r is an irrational number, we can write it as r=n+s, where s is an irrational number in the interval (0,1) and n is an integer. That means that r-s=nis a rational number, which in turn means r and s are in the same equivalence class, i.e. [r] = [s]. That means the elements of R/Q look like {0} U {a set of irrationals in the interval (0,1)}. But what set of irrationals, exactly? Not all of them; for example take the decimal portion of PI (=0.14159...) and add 0.5 to it to get 0.64159.... Both are irrational numbers yet their difference is ½, so they generate the same coset, in other words, they are collapsed to the same element of R/Q. On the other hand, it is known that sqrt(2), sqrt(3) and sqrt(2)-sqrt(3) are all irrational numbers. That means the cosets [sqrt(2)] and [sqrt(3)] are distinct elements of R/Q. So, some irrationals collapse to the same element in R/Q but not all do. So the question becomes: is there a way to choose or describe a set of irrational numbers that represent the distinct non-zero cosets of R/Q? The Axiom of Choice implies that, yes, one can choose a set of irrational numbers in the interval (0,1) that form a distinct complete set of cosets for R/Q. The catch, however, is that the Axiom of Choice gives no recipe for how to choose or describe such a set of representatives. What we can say is that two non-zero elements of R/Q, call them [r] and [s], are equivalent if and only if their decimal representations differ by only a finite number of digits. So a non-zero coset consists of all the irrational numbers in (0,1) that differ from each other by only a finite number of digits. It would seem that surely there must be a way to methodically pick from each coset a "canonical" representative. A likely candidate might be to pick the smallest member in each coset, but of course that fails because there is no smallest member in each non-zero coset; same for largest. What one would ideally like is a choice function f:R/Q -> (0,1) such that for any two cosets C and D, f(C+D) = f( C)+f(D) mod(0,1). As far as I know, no such choice function has been described and, indeed I do not know if it is even possible to define such a function in the standard ZF language. Note that the AC does not imply that such a function as f, above, exists. It only says that a choice function exists, but says nothing about how it will behave arithmetically as described above.
(*)Note: The statement above that two cosets [r] and [s] are equal iff r and s differ by a finite number of digits is almost right, but ignores the possibility that r-s might be a repeating decimal, like 1/9. If we permit ourselves to use the repeating decimal symbology of a bar over the repeating segment of decimals, then the original statement stands true.