Consider the (inhomogeneous) minimal surface equation for functions $u,f:D\to \mathbb{R}$ for some smooth domain $D\subset \mathbb{R}^n$
$$Lu:=\operatorname{div} \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=f.$$
Then is it true that $u_1\leq u_2$ on $\mathbb{R}^n\setminus D$ and $Lu_1\geq Lu_2$ implies $u_1\leq u_2$ on $D$?
For the case that $D$ is bounded, as Leo Moos has noted, more details can be found in a textbook of minimal surface, cf. Lemma 1.26 in T. Colding, W. Minicozzi, A Course in Minimal Surfaces.
That is, let $F(X)=\frac{X}{\sqrt{1+|X|^2}},$ then \begin{equation} F(\nabla u_1)-F(\nabla u_2)=\left(\int_0^1dF\big(\nabla u_2+t(\nabla u_1-\nabla u_2)\big) dt\right)(\nabla u_1-\nabla u_2). \end{equation} From this, one can conclude that $v=u_1-u_2$ satisfies an equation of the form $$\operatorname{div}(a_{i,j}\nabla v)\leq 0,$$ where the matrix is defined as $(a_{i,j})=\int_0^1dF(\nabla u_2+t(\nabla u_1-\nabla u_2)) dt.$
In particular, for a unit vector $V$ and a vector $X$, we have $$dF(x)V=\frac{V}{\sqrt{1+|X|^2}}-\frac{\langle V,X\rangle}{(1+|X|^2)^{\frac{3}{2}}}X.$$ Thus, $$(1+|X|^2)^{\frac{3}{2}}\langle V, dF(X)V\rangle=(1+|X|^2)-\langle V,X\rangle^2\geq 1, $$ which means $(a_{i,j})$ is positive, therefore, the usual comparison principle gives the claim.
