Going from the topological index idea: let $C$ be a circle around one of the isolated zeros, and let $D$ be a disk containing $C$. By the trace theorem, your vector fields in $W^{1,2}(D)$ restricts to vector fields $W^{1/2,2}(C)$ on the circle. Since they do not vanish, you can regard them as vector fields $W^{1/2,2}(C, \mathbb{S}^1)$. (If there's any problem with my argument, it would be this step, passing from an $\mathbb{R}^2$ valued function to $\mathbb{S}^1$ valued one.)

The $W^{1/2,2}(C,\mathbb{S}^1)$ functions are continuously embedded in $VMO(C,\mathbb{S}^1)$ since $C$ is one dimensional, and one can use the $VMO$-degree theory (originally due to Boutet de Monvel and Gabber, and extended by Brezis and Nirenberg, see [this survey by Brezis](https://link.springer.com/chapter/10.1007/0-8176-4467-9_3)). The upshot is that for continuous functions the VMO degree coincides with the the usual topological degree, and the $VMO$-degree is continuous under $VMO$-convergence. So I think [Pietro's argument against the $W^{2,2}$ case using degree theory](https://mathoverflow.net/questions/346238/can-we-approximate-a-vector-field-on-the-plane-with-non-vanishing-vector-fields#comment866546_346238) should carry over also, telling you that you shouldn't be able to do your approximation.