The restriction of a continuous map $D^2\to S^1$ to $\partial D^2\to S^1$ must have degree zero. Is that statement true or false if the map is only $W^{1,2}(D^2;S^1)$ and continuous on $\partial D^2$?

In two dimensions, the Sobolev space $W^{1,2}$ is at the borderline regularity and does not embed into the space of continuous functions. However, we can ask that a map's values be on the circle almost everywhere, giving us the space $W^{1,2}(D^2;S^1)$. Such a map has a well-defined restriction to the boundary; the restriction is a $W^{1/2,2}(\partial D^2;S^1)$ map. If it so happens that this restriction is continuous, we can ask about its degree.

A standard example of a discontinuous $W^{1,2}$ function is $f=\log(\log(4/r))$, and we can use it to construct a discontinuous $W^{1,2}$ circle-valued map $g=e^{if}$. However, this map has degree zero on the boundary, and it's not clear to me if one can do a different construction to get a degree one map.

Another good example to consider is $e^{i\theta}$, which is a $W^{1,p}(D^2;S^1)$ map for any $p<2$, and has degree one on the boundary.