(1). No, there are functions $[0,1]^2\to R$ such that every point preimage has Hausdorff dimension $>1$.
Recall that there exist functions $g: R\to R$ such that the graph of $g$ has Hausdorff dimension (at every point) $>1$, for instance, Weierstrass nowhere differentiable function has this property (see here). Now, consider the foliation of $R^2$ by the curves $C_t=\{(x, g(x)+t): x\in R\}$. Then define the function
$$
f: R^2\to R
$$
sending each $p\in R^2$ to $t\in R$ such that $p\in C_t$. This function is clearly continuous. Thus, we obtained a continuous function $f: R^2\to R$ such that for every $t\in R$, $f^{-1}(t)$ has Hausdorff dimension $>1$.
(2). Edit. Now, to your second questions. An easy estimate is that
$$
\sup_a H^1(f^{-1}(a))\ge \sqrt{2\pi}-2 \approx 0.506.
$$
The obvious conjecture, of course, is that this supremum is $\ge 1$, but I do not know how to prove it; you may have to play with quantities like Cheeger constant and lowest eigenvalue of the Laplacian to see if you can improve on my bound. Here is a proof of the $0.5$ estimate. Set $Q=[0,1]^2$.
For each $t\in {\mathbb R}$, define subsets
$$
E_t=\{p\in Q: f(p)\le t\}, E^c_t=Q\setminus E_t.
$$
Let $A_t$ denote the area of $E_t$. Let $L_t$ denote the 1-dimensional measure of the boundary of $E_t$. It equals the sum of $\ell_t=H^1(f^{-1}(a)$ and the length $m_t$ of $E_t\cap \partial Q$. Similarly, define
$$
L_t^c= H^1(\partial E_a^c)= \ell_t + (4-m_t).
$$
From the Euclidean isoperimetric inequality we obtain
$$
L_t\ge \sqrt{4\pi A_t}, L_t^c\ge \sqrt{4\pi (1-A_t)}.
$$
By adding the inequalities together, we get:
$$
\ell_t\ge \sqrt{\pi}(\sqrt{A_t} + \sqrt{1-A_t}) -2.
$$
Since $A_t$ is a continuous function of $t$, there exists $t$ such that $A_t=1/2$; for this $t$ we obtain
$$
\ell_t \ge \sqrt{2\pi}-2.
$$
(3) As for your last question:
Lemma. If $f: [0,1]^2\to R$, then there exists $t\in R$ such that $f^{-1}(t)$ has Hausdorff dimension $\ge 1$.
Proof. Indeed, suppose that there exists $f$ such that $f^{-1}(t)$ has Hausdorff dimension $< 1$ for all $t\in R$. Then, since Hausdorff dimension is always $\ge $ Lebesgue covering dimension (see e.g. "Dimension theory" by Hurewicz and Wallman) it follows that all subsets $f^{-1}(t)$ have covering dimension $0$. However, every point separates ${\mathbb R}$. Therefore, by taking point preimages, we see that there will be zero-dimensional subsets separating the unit square. This is impossible (a proof can be found again in "Dimension theory", by Hurewicz and Wallman, page 48). qed
