This answer tries to walk a fine line - I think this is a very natural and simple-to-pose question which nonetheless benefits from quite technical material, and that makes exposition difficult. I've attempted to keep things readable and not too misleading, but please let me know if I can improve this!
First of all, there is a very precise and elementary sense in which most sets of reals are non-measurable if we use a category-, rather than measure-, theoretic approach: in the natural topology on equivalence classes of sets of reals, where the equivalence relation is "differs by a null set," the collection of (classes of) measurable sets is meager. See here. However, I'm not sure how satisfying this ultimately is, so let me try for something more general here.
There is a surprisingly useful conceptual shift we can employ (if we're willing to engage in a lot of blackboxing): instead of talking about "mostness," talk about "genericity!" "Most $X$ are $Y$" is of similar flavor to "A generic $X$ is $Y$," and the latter lets us shift from trying to measure sets of things to thinking about approximating individual things. (Admittedly this is actually just retreating to the category-focused approach, but at a level of generality that lets us subsume measure too.) There is a very general, if quite technical, method provided by set theory for talking about "generic" objects - in this case, generic sets of real numbers - called forcing. This will let us state and prove, in a precise sense, things like "Every 'non-silly' type of genericity has the property that generic sets of reals are non-measurable." The cost we pay in technical difficulty is in my opinion justified by the resulting generality.
Very roughly speaking, suppose we have a partial order $\mathbb{P}$ whose elements (called "conditions") we think of as approximations to a hypothetical set of reals $G$. Then we can ask what sort of properties this $G$ will have if it is sufficiently generic. The exact meaning of genericity comes from the choice of $\mathbb{P}$ in a somewhat subtle way (this generalizes how measure and category each make sense as notions of size, but there are sets which are "big" in one sense but "small" in the other), so to start with I'll just go with the simplest choice of $\mathbb{P}$ that works nicely here:
Let $\mathbb{P}$ be the set of pairs $(\mathsf{In},\mathsf{Out})$ of disjoint countable sets of reals.
The idea is that a given condition $p=(\mathsf{In}_p,\mathsf{Out}_p)$ consists of countably many positive facts (each $x\in\mathsf{In}_p$ should be in the $G$ we're imagining building) and countably many negative facts (ditto but dually). This tries to capture the idea that individual real numbers are being put into, or kept out of, $G$ independently; we're "flipping continuum-many coins."
Of course, there is the annoying issue of the countability requirement. This is for technical reasons. Roughly speaking, if we allowed the decision-making sets to be "too big" (= size continuum) then a single condition could determine the whole set all at once and so $\mathbb{P}$ wouldn't be doing anything, and if we required the sets to be "too small" (= finite) then ... well, actually the disaster in this case is a bit hard to describe, but trust me that it would be annoying. Of course these two issues still leave lots of room for variation, but in my opinion the $\mathbb{P}$ defined above is indeed the most natural choice.
It now turns out that the notion of genericity we get from $\mathbb{P}$ does in fact imply non-measurability.$^1$ So, if we believe that $\mathbb{P}$ describes the "right" way of approximating a hypothetical set of reals, then it is in fact the case that a "generic" set of reals is non-measurable.
But maybe you're not convinced: what if we used a different $\mathbb{P}$, and consequently a different notion of "genericity," in posing this question? This is where the rather general line above comes into play: as it turns out,$^2$ the specific $\mathbb{P}$ above almost didn't matter at all. All we needed was for $\mathbb{P}$ to satisfy a couple basic properties, namely:
$\mathbb{P}$ "adds no new reals" - this is a technical property, but it's more than enough to have $\mathbb{P}$ be countably closed in the sense that for any countable sequence of increasingly-strong conditions, there is a single condition which is at least as strong as all of them (for the $\mathbb{P}$ above this is a consequence of the fact that $\aleph_0\times\aleph_0=\aleph_0$).
No condition in $\mathbb{P}$ pins down the set of reals up to a measure-zero set.
(This is an immediate consequence of $(i)$ the forcing theorems, $(ii)$ the fact that every null set is contained in a null $F_\sigma$ set, $(iii)$ the fact that Borel sets are coded by reals, and $(iv)$ the fact that a set is measurable iff it differs from a Borel set by a null set.)
So in a precise sense, "a generic set of reals is non-measurable" is true for any reasonably-nice notion of genericity.
$^1$One way to make this somewhat precise is the following: if $M$ is a "toy model" of set theory, then any time we "slightly expand" $M$ by adding a set of reals $G$ which is generic in the sense of $\mathbb{P}$ the resulting model $N$ thinks that $G$ is non-measurable. Of course it doesn't make sense to talk about adding something to the actual universe of sets (without doing a lot of subtle circumlocution, anyways), but this has a vague affinity with genericity ideas coming from algebraic geometry - think of $M$ as the part of a variety seen by a very small field, for example.
$^2$A friend asked me to add some detail here, so I'll do so - but at this point the answer becomes unavoidably technical (and going forward I'll use the standard terminology/notation of forcing without comment). The proof basically amounts to cheating with Borel codes. Suppose $M$ is our ground model and $\mathbb{P}\in M$ is a forcing notion which adds no new reals, and $\nu$ is a $\mathbb{P}$-name for a set of reals. Let $p\in\mathbb{P}$ be a condition forcing that $\nu$ is measurable. If $G\ni p$ is $\mathbb{P}$-generic over $M$, then $M[G]\models$ "The set $\nu[G]$ differs from a Borel set by a null set." But Borel sets in $M[G]$ are coded by reals in $M[G]$, and $\mathbb{P}$ adds no new reals, so there is a Borel code $b\in M$ which when evaluated in $M[G]$ is within a null set of $\nu[G]$. But again since $\mathbb{P}$ adds no new reals, the Borel-evaluation of $b$ in $M$ is the same as the evaluation of $b$ in $M[G]$. Putting this together, there is a set $B\in M$ which ($M$ thinks) is Borel and satisfies $p\Vdash_\mathbb{P}m(\nu\Delta B)=0$ (where $m$ and $\Delta$ denote Lebesgue measure and symmetric difference respectively).
