What is the moduli space of germs of one-sided complex structures near the circle? Consider a one-sided ( say, internal) neighborhood $U$ of the unit circle $S$ ( $U$ contains $S$) on the plane with a choice of smooth complex stricture $\tau$ on $U$.
By smoothness of $\tau$ on $U$ we mean that corresponding operator of almost-complex structure $J$( defined in the interior of $U$) as a matrix with respect to the coordinate basis on the plane, has smooth extension to $S$.
Two such pairs $(U_1,\tau_1)$ and $(U_2,\tau_2)$ are equivalent if there are subneighborhoods $V_1\subset U_1$ and $V_2\subset U_2$ ( both containing $S$) and a diffeomorphism $f: V_1\rightarrow V_2$ which is biholomorphic in the interior of $V_1$ with respect to $\tau_1, \tau_2$.
What is the moduli space of such structures?
 A: 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 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.
A: I will assume you mean that $U$ is a "deleted" neighborhood, i.e. that it does not contain $S$, since that is the more general assumption. Then there are two equivalence classes: one represented by the standard complex structure on the open disc $D$ bounded by $S$ and the other, the ``cusp'', represented by a neighborhood of $\infty$ in $\mathbb{C}$. The proof uses that every complex structure on the annulus is biholomorphic to one of $\mathbb{C}-0$, $D-0$, or $D - \overline{r D}$ for some $r \in (0,1)$.
