I believe I have an answer to my own question: the boundary map indeed must have degree zero, though I would still be curious about simpler proof methods that might more easily generalize to higher dimensions.
The trick is that we can compute the degree of a map $g\colon\partial D^2\to S^1$ by computing $\int_{\partial D^2}\bar g\,dg$. This integral will be $2\pi i$ times the degree. But, moreover, this expression is well-defined for $g\in W^{1/2,2}(\partial D^2)$ because $\bar g\in W^{1/2,2}(\partial D^2)$ and $dg\in W^{-1/2,2}(\partial D^2)$. In other words, the notion of the degree of a map $\partial D^2\to S^1$ can be extended continuously from $C^\infty(\partial D^2;S^1)$ to $W^{1/2,2}(\partial D^2;S^1)$.
However, that's not the end of the story, because functions in $W^{1,2}(D^2;S^1)$ are limits of functions in $C^\infty(D^2;\mathbb C)$ that converge pointwise almost everywhere to $S^1$, and it is a nontrivial fact to show that they are actually limits of functions in $C^\infty(D^2;S^1)$. (Consider $e^{i\theta}$, which is in $W^{1,p}(D^2;S^1)$ for $p<2$, but to view it as a $W^{1,p}$-limit of smooth functions, we need to cut it off to zero near the origin.) Fortunately, Schoen and Uhlenbeck provide a proof in section 4 of Boundary regularity and the Dirichlet problem for harmonic maps.