The minimal surface equation is uniformly elliptic, at least in the sense that its linearization at any solution is uniformly elliptic. It will be convenient to rewrite the equation as 
$$\nabla \cdot \left(\frac{\nabla u}{\sqrt{1 + |\nabla u|^2}}\right) = 0$$
(this is the standard way that the equation is written on euclidean space, anyways). We now have the coefficient $A_{ij} = (1 + |\nabla u|^2)^{-1/2} \delta_{ij}$, and the PDE is $Pu := \nabla \cdot (A\nabla u) = 0$.

By elliptic regularity, the solution $u$ is smooth, so there exists $\varepsilon > 0$ such that on $B_{3/4}$ (say) we have $|\nabla u| < \varepsilon^{-1}$. On $\{|\nabla u| < 1\} \cap B_{3/4}$ we have $A_{ii} > 1/10$, and on $\{|\nabla u| \geq 1\} \cap B_{3/4}$ we have $A_{ii} > \varepsilon^{1/2}/10$. Therefore $P$ is a uniformly elliptic operator and we can apply the usual proof of Harnack's inequality.