There are essentially two cases when the limit is not a covering map: (a) the limit is constant, (b) the images $S_n:=f_n(D)$ are annuli which collapse to a circle. There could be a more elementary explanation, but the one below uses Fuchsian groups and a bit of hyperbolic geometry.

Sketch of the proof. Each surface $S_n$ can be identified with the quotient $D/G_n$, where $G_n$ is a Fuchsian group (a discrete torsion-free isometry group of the unit disk $D$, which we will regard as the hyperbolic plane). Thus, I will think of $S_n$ as complete hyperbolic surfaces. Now, the sequence of discrete groups $G_n$ will subconverge in "Chabauty topology" (or "geometrically" in Thurston's terminology) to a closed subgroup $G$ of $PSL(2,R)$, the isometry group of $D$. Most the the time the group $G$ will be discrete and torsion-free. The one exception is when the groups $G_n$ converging to $G$ are elementary, i.e., cyclic, then the limit could be 1-dimensional (this is where you discover the annular example). Suppose now that the group $G$ is discrete and consider the quotient surface $D/G=S$. Then (after you choose base-points) the surfaces $S_n$ converge to $S$ in "Gromov-Hausdorff topology." (You may have to read something like "Lectures on Hyperbolic Geometry" by Benedetti and Petronio if the terminology is unfamiliar.) This means that large metric balls in $S$ (in the hyperbolic metric of $S$) are nearly conformal to corresponding metric balls in $S_n$ (i.e., quasiconformal with small quasiconformality constants). Thus, your limiting holomorphic map $f: D\to D$ (which I assume to be nonconstant) would factor as a composition of $p: D\to S$ (the universal covering map) and a holomorphic map $F: S\to D$, which I claim is 1-1. If the map $F$ were the limit of a sequence of conformal maps $S\to D$, you would be done (say, by Rauche's theorem), as you observed in your question. Instead it is a limit of almost conformal quasiconformal maps above ($S\to S_n$) defined on larger and larger compact subsets of $S$. Now, you have to check that the same conformal arguments that you know would go through in the quasiconformal situation as well (since they are nothing but degree arguments and work for local homeomorpisms as well), which concludes the proof.

Edit: Here is a better and more general argument which works in (smooth) topological setting:

Theorem. Let $M, N$ be smooth manifolds of the same dimension, so that $M$ is connected and $N$ is compact. Let $f_i: M\to N_i\subset N$ be a sequence of (locally diffeomorphic) covering maps, which converges on compacts in $C^1$-topology to a local diffeomorphism $f_0: M\to N$, whose image is a domain $N_0\subset N$. Then $f_0: M\to N_0$ is also a covering map.

Proof. Let $g$ be a Riemannian metric on $N$. Define function $\rho_i$ on $N_i$ to be the (minimal) distance function from $x\in N_i$ to the frontier of $N_i$ in $N$, $i\ge 0$. Then the sequence of functions $\rho_i$ is semicontinuous in the sense that for every $x\in N_0$, $\lim inf \rho_i(x) \ge \rho_0(x)$. Now define (Lipschitz) continuous Riemannian metrics $g_i=g/\rho_i$ on $N_i$. These are my replacements of Poincare metrics. used in the original argument One verifies easily that every $g_i$ is complete (here and below completeness is in Cauchy sense since exponential map need not be defined even locally).

It is a standard fact of Riemannian geometry that a local diffeomorphism of smooth connected manifolds is a covering map iff the pull-back of some (every) complete Riemannian metric on the target is complete as a Riemannian metric on the domain. It is easy to check that the proof goes through in the case of continuous Riemannian metrics.

Now, the continuous Riemannian metric $\tilde{g}_i=f_i^*(g_i)$ is complete for every $i>0$. On the other hand, for every $i$,
$$
f_i^*(g_i)\le f_0^*(g_0).
$$
Thus, completeness of the metrics $\tilde{g}_i$ for $i>0$ implies completeness of $\tilde{g}_0$. Hence, $f_0: M\to N_0$ is a covering map. Qed.