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 $C^\infty(D^2;S^1)$, whose boundary restrictions indeed have degree zero. (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.
In higher dimensions, it appears that this smooth approximation issue has also been resolved. See the introduction of this recent paper, for example. To generalize the above argument to higher dimensions, 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*} \frac1{3!\cdot\operatorname{vol}(\partial B^4)}\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 $g\in W^{1,4}(B^4;\mathbb R^4)$.
See also the introduction of this Brezis and Nirenberg paper.