Since $F$ is bicomplex-holomorphic, we necessarily have $\frac{\partial F}{\partial Z^\dagger} = 0$, and your condition, let's call it the bicomplex Beltrami equation, $$ \frac{\partial F}{\partial Z^\dagger}=\mu(Z)\frac{\partial F}{\partial Z} $$ implies that either $\mu$ is the zero measure or $F$ is constant.
By the way, any bicomplex-holomorphic function $F(Z)$ becomes - after a linear change of coordinates - of the form $f_1(w_1) \oplus f_2(w_2)$ for some $\mathbb{C}$-holomorphic functions $f_1$ and $f_2$.
A more appropriate generalization of quasi-conformality is to only require $F: U \to \mathbb{BC}$, $U \subseteq \mathbb{C}^2$ open, to be complex-differentiable in the usual sense as a function of several complex variables or even only real-differentiable as a function of 4 real variables. After all, in the Beltrami equation for $\mathbb{C}$, $f$ is only real-differentiable, which is what makes the problem interesting.