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: I guess this doesn't answer the question of whether we can remove the dependence of $C$ on the ellipticity constant. The answer is no, $C$ must depend on the ellipticity. I don't have an explicit example using a minimal surface, but you can already see why this is problematic by taking $(au')' = 0$ on $(0, 1)$. Clearly this is equivalent to the PDE $$au'' + a'u' = 0$$ and we can take $a(x) = x^2 + \delta$. Then the solution to the equation is $$u(x) = \delta^{-1/2} \arctan(\delta^{-1/2} x)$$ which is positive on $(0, 1)$, and for $\delta$ small is approximately arbitrarily well by $x/\delta$. Thus the Harnack inequality, taken say in $(1/4, 3/4)$ becomes arbitrarily bad!

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 $\varepsilon \to 0$.

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.

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: I guess this doesn't answer the question of whether we can remove the dependence of $C$ on the ellipticity constant. The answer is no, $C$ must depend on the ellipticity. I don't have an explicit example using a minimal surface, but you can already see why this is problematic by taking $(au')' = 0$ on $(0, 1)$. Clearly this is equivalent to the PDE $$au'' + a'u' = 0$$ and we can take $a(x) = x^2 + \delta$. Then the solution to the equation is $$u(x) = \delta^{-1/2} \arctan(\delta^{-1/2} x)$$ which is positive on $(0, 1)$, and for $\delta$ small is approximately arbitrarily well by $x/\delta$. Thus the Harnack inequality, taken say in $(1/4, 3/4)$ becomes arbitrarily bad!