When is a harmonic map biholomorphic? Let $D_1$ and $D_2$ be simply-connected bounded open domains on $\mathbb{C}$. Riemann mappping theorem tells us that there exist biholomorphisms between them.
On the other hand, let $\gamma : \partial D_1 \rightarrow \partial D_2$ be a $C^2$ diffeomorphism. We know that there exists a unique harmonic map $f_{\gamma} : \bar{D_1} \rightarrow \mathbb{C}$ such that $f_{\gamma}|_{\partial D_1} = \gamma$. However, $f_{\gamma}$ may not be holomorphic. It is even not guaranteed that $f_{\gamma}$ is injective, or $f_{\gamma}(\bar{D_1})=\bar{D_2}$.
Here comes the question. What is the necessary and sufficient condition imposed on $\gamma$, so that $f_{\gamma}$ is a biholomorphism from $D_1$ onto $D_2$?
 A: Radó–Kneser–Choquet theorem gives a nice sufficient condition for $f_\gamma$ to be a diffeomorphism of $D_1$ onto $D_2$. (Perhaps you already know that, but did not mention it.) The theorem was recently generalized to some extent: see Invertible harmonic mappings, beyond Kneser by Alessandrini and Nesi, and subsequent papers by Kalaj. The conditions for invertibility in these papers are not entirely geometric like the convexity condition; they involve the map $\gamma$ itself. 
But you ask for $f_\gamma$ to be a conformal diffeomorphism between the domains. The general answer to that is tautological: $f_\gamma$ has to be the boundary values of a conformal map. Consider that the space of conformal maps $D_1\to D_2$ is described by just three real parameters. So the maps you are interested in form a three-dimensional subset of the huge space of diffeomorphisms $\partial D_1\to\partial D_2$. The conformality of the harmonic extension is too fragile a property to have an analytic  description. 
So, if you want to check whether some particular extension is conformal, you will probably end up differentiating the Poisson integral and checking whether its $\partial/\partial \bar z$ derivative vanishes. 
However in the special case when $D_1$ is the unit disk, you can use the following fact: the Poisson extension of $\gamma$ is holomorphic if and only if 
$$\int_{\partial D_1} z^n \gamma(z)\,dz =0,\qquad n=1,2,\dots \tag1$$
Indeed, (1) and the fact that $(z^n)$ are orthogonal on the unit circle say  that $\gamma(z) = \sum_{k=0}^\infty c_k z^k$ on the boundary. This same formula provides the holomorphic extension of $\gamma$. Moreover, this extension is indeed a diffeomorphism, because $\gamma$ winds around every point of $D_2$ once, allowing us to obtain injectivity from the argument principle.
