Suppose that we have an embedding $f:\mathbb{R}^2\to\mathbb{R}^3$ which belongs in the Sobolev space $W^{1,p}_{loc}(\mathbb{R^2},\mathbb{R}^3)$ for some $p>2$. Is it true then that for any two points in $f(R^2)$ there exists a curve of finite length connecting them?
Fuglede's theorem: Suppose that $f:\mathbb{R}^n\to\mathbb{R^m}$ is in $W^{1,p}_{loc}$ for any $p$. Then the family of paths which have a closed subpath on which $f$ is not absolutely continuous has $p$-modulus zero.
Added: The modulus of a path family $\Gamma$ is defined to be $M_p(\Gamma)=\inf_{\rho}\int_{\mathbb{R}^2}\rho^pdm$, where the infimum is taken over all Borel functions $\rho:\mathbb{R}^2\to[0,\infty]$ such that $\int_\gamma\rho ds\geq 1$ for all locally rectifiable path $\gamma\in \Gamma$.
PARTIAL PROGRESS. As you mention in comments (and you should really edit the question to include this important piece of information), by Fuglede's theorem $f(\gamma(t))$ is absolutely continuous for almost all path $\gamma\colon [0, 1]\to \mathbb R^2$. In particular, $f(\gamma(t))$ is finite length for almost all paths.
Now take $p=f(x)$ and $q=f(y)$. Let $$\Gamma=\{\gamma\colon [0, 1]\to \mathbb R^2\ :\ \gamma(0)=x, \gamma(1)=y\}.$$ For each $\gamma\in\Gamma$, the path $f(\gamma(t))$ connects $p$ and $q$. The difficulty is that $f(\gamma(t))$ may a priori have infinite length.
So aiming for a contradiction suppose that $f(\gamma(t))$ has infinite length for all $\gamma\in \Gamma$. I do not have the necessary definitions, because you did not explain them, so I cannot continue. (What does "almost all path" mean, exactly? What measure you put on the space of all paths?).
However, my gut feeling is that this must be in contradiction with that Fuglede's theorem mentioned above. For, whatever that "almost all path" means, I doubt that $\Gamma$ has measure zero. Therefore there must be at least one path $\gamma$ in $\Gamma$ such that $f(\gamma(t))$ is absolutely continuous.