I've been sitting down can't quite tell if this is true or not, but I suspect that it should be.
Edit: Suppose that $\Omega$ is a open, connected subset of $\mathbb{C}$, and suppose that $(f_n) \subset H(\Omega)$ and $f_n$ converges to $f \in H(\Omega)$ on compact subsets of $\Omega$; if $g$ is a holomorphic self-map of $\Omega$, then is is true that $(f_n \circ g)$ converges uniformly on compact subsets of $\Omega$ to $f \circ g$?
Addendum: Here's my attempt at a solution.
Fix any compact subset $K$ of $\Omega$ and any $\epsilon > 0$. There are two cases to consider: when $g$ is constant and when $g$ is nonconstant. In the second case, the Open Mapping Theorem gives us that $g(\Omega) \subset \Omega$ is an open set. By the continuity of $g$, the image of $K$ under $g$, $g(K)$, is a compact subset of $g(\Omega)$, and hence, a compact subset of $\Omega$. Thus, by the assumptions made on the sequence $(f_n)$, there exists some positive integer $n_0$ for which $$ \max_{w \in g(K)} |f_n(w)-f(w)| <\epsilon $$ whenever $n \geq n_0$. s, it follows that $f_n \circ g$ converges uniformly on compact subsets of $\Omega$ when $g$ is nonconstant.
In the case when $g$ is constant, the image $g(K)$ is still a compact subset of $\Omega$ and we apply a similar argument.
Does this seem correct?