@IanAgol: If I'm understanding you correctly, $j \circ z^{1/3}$ cannot be analytically continued to any point where $j$ has a simple zero (and aren't all zeros of $j$ simple?). So in that case, what you'll get for $M$ is a $3$-fold cover of $\mathbb{H}^2$ minus the zeros of $j$, and the map from $M$ to the complex plane is that 3-fold covering.

(another example: if you do this to a germ of $z^{1/3}$ around $1$, then you'll get the usual 3-fold cover from $M \cong \mathbb{C} \setminus 0$ to itself)

(note e.g. that there is a component $M$ that is the maximal analytic continuation of the entire function $f(z) = z^3$, and the map $p\colon M \rightarrow \mathbb{C}$ is a homeomorphism)

@IanAgol: I think you're thinking of something slightly different, namely whether the holomorphic map itself is a covering space. Essentially by definition the map I described is a local homeomorphism (note that it outputs a point in the domain of the function, not its range).

@TomGoodwillie: That's kind of what I expected. Do you know a counterexample?