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.

EDIT 2: Here's an explicit example. Recall that the catenoid is the surface of revolution corresponding to the catenary
$$z = f(x) := \varepsilon \cosh\left(\frac{x}{\varepsilon}\right)$$
and it is minimal. Taking only half of the surface of revolution, we get a minimal graph $z = u(x, y)$ with $u > 0$. Consider the restriction of $u$ to the ball $B$ of radius $1/4$ centered on $(x, y) = (1, 0)$. Then the graph of $u$ contains the the catenary, so $\inf_B u \leq f(0.75)$ and $\sup_B u \geq f(1.25)$. Now we estimate the ratio
$$\frac{\sup_B u}{\inf_B u} \geq \frac{f(1.25)}{f(0.75)} = \frac{\cosh(1.25/\varepsilon)}{\cosh(0.75/\varepsilon)}$$
which blows up as $\varepsilon \to 0$.