I think that, even if you include the boundary $S$, they are all equivalent. In saying so, I'm assuming that the final 'it' in your first sentence refers to the plane, not to $U$ and that you are considering smooth complex structures.
The reason is this: Take a sufficiently small open neighborhood $W$ of the closure of the disk $\Delta$ that $S$ bounds that is also a disk. By the Riemann mapping theorem, there is a biholomorphism of $W$ (endowed with the given complex structure) with the unit disk $\Delta$ endowed with its standard complex structure.
The image $S'\subset \Delta$ of $S$ under that biholomorphism is a smoothly embedded curve that bounds a disk $D'\subset\Delta$. By the known boundary regularity in the Riemann mapping theorem, there is a map $(D',S')\to (\Delta,S)$ that is a biholomorphism in the interior and smooth up to and including the boundary (see the discussion at Riemann mapping theorem and smoothness on the boundaryRiemann mapping theorem and smoothness on the boundary). Composing these two biholomorphisms gives the equivalence of the original $(U,S)$ with the given complex structure with a $(U',S)$ for which the complex structure is the standard one.
Now pass to germs, and you have the equivalence that you wanted.
