The following is too long for a comment. I will suppose here that your set of measure zero is compact.
In this case, Your question is closely related to so-called continuous analytic capacity. Let $K$ be a compact set in the plane, and let $\Omega$ be the complement of $K$ with respect to $\mathbb{C}_\infty$. The continuous analytic capacity $\alpha(K)$ is defined as
$$\alpha(K):= \sup \{ |f'(\infty)| : f \in A(\Omega), \|f\|_{\infty} \leq 1 \},$$
where $A(\Omega)$ is the set of all functions holomorphic in $\Omega$ that extend continuously to $\mathbb{C}_\infty$.
It is not difficult to prove that $$\alpha(K)=0$$ if and only if for every open set $U$ with $K \subseteq U$, every $f$ holomorphic on $U \setminus K$, continuous on $U$, extend analytically to the whole of $U$.
So what you're looking for is a characterization of the compact sets $K$ with $\alpha(K)=0$. This is certainly very difficult, since it took more than a hundred years to solve a similar problem (but with analytic capacity $\gamma$ instead of $\alpha$), the so-called Painlevé's problem.
It is easy to prove however that if $\alpha(K)=0$, then the area of $K$ must be zero. You're asking if this condition is sufficient. Most likely it is not, but I don't have a counterexample right now.. Usually, for problems related to removable sets for holomorphic functions, it is more useful to consider hausdorff measure instead.
Anyway, I suggest you look in Garnett's book "Analytic Capacity and measure". See also the recent book by Dudziak, "Vitushkin's conjecture for removable sets."
EDIT (Answer to the question)
Every compact set $K$ with Hausdorff dimension strictly between $1$ and $2$ is a counterexample. An easy way to see this is to use Frostman's lemma. Indeed, using this lemma, there exists a nonzero Borel measure $\mu$ supported on $K$ with growth $\mu(B(z,r)) \leq C r^{1+\epsilon}$. The growth of $\mu$ implies that the Cauchy Transform
$$f(z):=\int_{K} \frac{1}{z-\zeta}d\mu(\zeta)$$
is continuous on $\mathbb{C}$ and holomorphic on $\mathbb{C} \setminus K$. Furthermore, $f(z)\rightarrow 0$ as $z \rightarrow \infty$ and so in particular, $f$ is bounded on $\mathbb{C}$. However, $f$ is non-constant because $zf(z) \rightarrow \mu(K) \neq 0$. Therefore $f$ does not extend analytically to all of $\mathbb{C}$, for otherwise it would be constant by Liouville's Theorem.
