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 other proof methods.

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](https://projecteuclid.org/download/pdf_1/euclid.jdg/1214437663).

In higher dimensions, it appears that this density issue has also been resolved. See the introduction of [this recent paper](http://link.springer.com/article/10.1007/s11118-016-9558-x), for example. It remains to express the degree as an integral. Intuitively, we want to integrate the determinant of the Jacobian. For maps $g\colon\partial B^4\to S^3$, for example, the expression is
\begin{equation*}
\int_{\partial B^4}g\wedge dg\wedge dg\wedge dg.
\end{equation*}
Explaining the notation, we view $g$ as a $\mathbb R^4$-valued zero-form on $\partial B^4$, and, correspondingly, $dg$ as a $\mathbb R^4$-valued one-form on $\partial B^4$. The $\wedge$ operation in this case means taking the usual wedge of forms on $\partial B^4$, along with taking the wedge in $\bigwedge^*\mathbb R^4$. (In particular, the operation is symmetric on $\mathbb R^4$-valued one-forms.) The resulting expression is a three-form on $\partial B^4$ with values in $\bigwedge^4\mathbb R^4\cong\mathbb R$.

Naively, because of borderline and negative regularity issues, this integral is not defined for $g\in W^{3/4,4}(\partial B^4;S^3)$. However, we can extend $g$ to a function in $W^{1,4}(B^4;\mathbb R^4)$ (not necessarily sphere-valued) so that the extension depends continuously on the boundary value. By Stokes' theorem, we see that
\begin{equation*}
\int_{\partial B^4}g\wedge dg\wedge dg\wedge dg=\int_{B^4}dg\wedge dg\wedge dg\wedge dg.
\end{equation*}
The right-hand side is indeed well-defined for $g\in W^{1,4}(B^4;\mathbb R^4)$.