Here is a shot in the dark (Disclosure: I really know nothing about this problem).

Let $G:=\mathsf{SU}(2)$ act on $G^3$ by simultaneous conjugation; namely, $$g\cdot(a,b,c)=(gag^{-1},gbg^{-1},gcg^{-1}).$$ Then the quotient space is homeomorphic to $S^6$ (see Bratholdt-Cooper).

The evaluation map shows that the character variety $\mathfrak{X}:=\mathrm{Hom}(\pi_1(\Sigma),G)/G$ is homeomorphic to $G^3/G,$ where $\Sigma$ is an elliptic curve with two punctures.

Fixing generic conjugation classes around the punctures, by results of Mehta and Seshadri (Math. Ann. 248, 1980), gives the moduli space of fixed determinant rank 2 degree 0 parabolic vector bundles over $\Sigma$ (where we now think of the punctures are marked points with parabolic structure). In particular, these subspaces are projective varieties.

Letting the boundary data vary over all possibilities gives a foliation of $\mathfrak{X}\cong G^3/G\cong S^6$. Therefore, we have a foliation of $S^6$ where generic leaves are projective varieties; in particular, complex.

Moreover, the leaves are symplectic given by Goldman's 2-form; making them Kähler (generically). The symplectic structures on the leaves globalize to a Poisson structure on all of $\mathfrak{X}$.

Is it possible that the complex structures on the generic leaves also globalize?

Here are some issues:

1. As far as I know, the existence of complex structures on the leaves is generic. It is known to exist exactly when there is a correspondence to a moduli space of parabolic bundles. This happens for most, but perhaps not all, conjugation classes around the punctures (or marked points). So I would first want to show that all the leaves of this foliation do in fact admit a complex structure. Given how explicit this construction is, if it is true, it may be possible to establish it by brute force.
2. Assuming item 1., then one needs to show that the structures on the leaves globalize to a complex structure on all of $\mathfrak{X}$. Given that in this setting, the foliation is given by the fibers of the map: $\mathfrak{X}\to [-2,2]\times [-2,2]$ by $[\rho]\mapsto (\mathrm{Tr}(\rho(c_1)),\mathrm{Tr}(\rho(c_2)))$ with respect to a presentation $\pi_1(\Sigma)=\langle a,b,c_1,c_2\ |\ aba^{-1}b^{-1}c_1c_2=1\rangle$, it seems conceivable that the structures on the leaves might be compatible.
3. Moreover, $\mathfrak{X}$ is not a smooth manifold. It is singular despite being homeomorphic to $S^6$. So lastly, one would have to argue that everything in play (leaves, total space and complex structure) can by "smoothed out" in a compatible fashion. This to me seems like the hardest part, if 1. and 2. are even true.

Anyway, it is a shot in the dark, probably this is not possible...just the first thing I thought of when I read the question.