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, I believe we're not quite done, 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 $S^1$, which might not necessarily be 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.)
What we can do is write down that, for smooth functions $D^2\to \mathbb C$, we have \begin{equation*} \int_{\partial D^2}\bar g\,dg=\int_{D^2}d(\bar g\,dg)=\int_{D^2}d\bar g\wedge dg. \end{equation*} The left-hand side and the right-hand side are continuous in $g\in W^{1,2}(D^2;\mathbb C)$, so the equality between them continues to hold for $g\in W^{1,2}(D^2;\mathbb C)$. The right-hand side is zero for $C^\infty(D^2;S^1)$ functions, but, again, these might not be dense in $W^{1,2}(D^2;S^1)$. However, by taking the conjugate, we see that $d\bar g\wedge dg$ is purely imaginary a.e. for any $g\in W^{1,2}(D^2;\mathbb C)$. We'll show that if $g$ is circle-valued, then $d\bar g\wedge dg$ is also real a.e., and hence zero.
We have that $g\bar g=1$ a.e., so $g\,d\bar g+\bar g\,dg=0$ a.e., since the left-hand side is well-defined in $L^1(D^2)$. By taking conjugates, we see that $g\,d\bar g$ and $\bar g\,dg$ are purely imaginary a.e.. Next, we have that, almost everywhere, \begin{equation*} d\bar g\wedge dg = g\bar g\,d\bar g\wedge dg = g\,d\bar g\wedge\bar g\,dg, \end{equation*} which is a.e. the product of two purely imaginary forms and hence real.