This is the so-called *conformal welding problem*. One can ask the same question for any Jordan curve $\gamma$ (non necessarily analytic). With this domain of definition, your map $\Gamma$ is well-known to be neither injective nor surjective. There are even orientation-preserving homeomorphisms of the circle analytic everywhere except at one point that are not the welding homeomorphism of any Jordan curve. However, the image of your map $\Gamma$ contains all *quasisymmetric* orientation-preserving homeomorphisms of the circle, this is sometimes referred to as the *fundamental theorem of conformal welding* and it was first proved by Pfluger in 1960. A simpler proof was later given by Lehto and Virtanen.

In general, it is difficult to explicitely reconstruct the curve $\gamma$ from the welding homeomorphism. For some welding homeomorphisms though, the associated curve $\gamma$ has a special form. For instance, if the homeomorphism is the $n$-th root of a Blaschke product of degree $n$, then the corresponding curve $\gamma$ is a proper polyonimial lemniscate of the same degree. Conversely, the welding homeomorphism of a proper polynomial lemniscate of degree $n$ is a $n$-th root of a Blaschke product. This was proved by Ebenfelt, Khavinson and Shapiro in the paper "Two-dimensional shapes and lemniscates", arXiv:1003.4567. See also arXiv:1406.3545 for a simpler proof and  a generalization to rational lemniscates.

If you're interested in conformal welding, a good survey is the one by Hamilton 
MR1966191 (2005e:30012) Hamilton, D. H.(1-MD) Conformal welding. Handbook of complex analysis: geometric function theory, Vol. 1, 137–146, North-Holland, Amsterdam, 2002. 30C35

**EDIT**
Perhaps I should add more details to what I mean exactly by the fact that the map $\Gamma$ is in general not injective. It is easy to see that if $T$ is a Möbius transformation, then $\gamma$ and $T(\gamma)$ have the same welding homeomorphism. The map $\Gamma$ is not injective *even modulo Möbius transformations*. The easiest way to see this is to consider a curve $\gamma$ of positive area and use the measurable Riemann mapping theorem to obtain an infinite-dimensional family of homeomorphisms of the sphere conformal outside $\gamma$. If $f$ is any such map, then it is easy to see that $\gamma$ and $f(\gamma)$ give rise to same welding homeomorphism. However, a dimension argument shows that the image of $\gamma$ under such a map $f$ cannot be always Möbius-equivalent to $\gamma$.

A sufficient condition for the uniqueness of the curve $\gamma$ from its welding homeomorphism is if $\gamma$ is *conformally removable*, i.e. if every  homeomorphism of the sphere conformal outside $\gamma$ is a Möbius transformations.