If you want to generalise the Reimannian mapping theorem to higher dimensions the boundary condition should be different. Namely, it is more reasonable to ask that the boundary of the disk is mapped to a submanifold $M$ of $\mathbb C^n$ or real dimension $n$ such that $TM\oplus J(TM)$ span $\mathbb C^n$ at each point of $M$ (such submanifolds are called totally real). In other words, you can not force the boundary of a holomorphic disk in $\mathbb C^2$ to be an arbitrary curve. The best that you can do is to chose a two-dimensional surface in $\mathbb C^2$ and then the disk will chose a curve on the surface that can be its boundary. Under this condition you can expect to get a finite dimension set of solutions. This is what Gromov is doing in his seminal paper
Pseudo-holomorphic curves in symplectic manifolds
Example. Let us consider the torus $\mathbb T^2$ in $\mathbb C^2$ given by $|z|=|w|$. Then obviously the line $z=w$ intersects $\mathbb T^2$ in a circle and we get a holomorphic disk with boundary on $\mathbb T^2$. Let us prove that the curves on $\mathbb T^2$ isotopic this circle and bounding a holomorphic disk in $\mathbb C^2$ form a family of real dimension $3$. Indded let $D^2$ be a holomorphic disk in $\mathbb C^2$ with boundary on $\mathbb T^2$ isotopic to the above circle. Consider the projections of $D^2$ to the lines $z=0$ and $w=0$. Since these projections are holomorphic it is not hard to see that they are one to one to maps to disks
$(|w|\le 1, z=0)$ and $(|z|\le 1, w=0)$. So $D^2$ is a graph of a holomorphic one to one map from one unit disk to the other. These maps (as we know well) form $SL(2,\mathbb R)$.