The statement is as follows. See page 30 of CM or below for an image of the proof.
If $u:\Omega\subset \mathbb{R}^2\to \mathbb{R}$ is a solution to the minimal surface equation, then for all nonnegative Lipschitz functions $\eta$ with support contained in $\Omega\times \mathbb{R}$,
$$\int_{\text{graph } u} |A|^2\eta^2\leq C\int_{\text{graph } u}|\nabla_{{\text{graph } u}}\eta|^2.$$
I understand the Lipschitz assumption, as that much regularity is sufficient for Stokes' theorem, and the logarithmic cutoff function they build in the next section is Lipschitz not differentiable. However, I'm not sure where the non-negative assumption is really used in the proof.
As far as I can tell, it's only used in the first inequality of (1.99), and one may just replace $\eta$ with $|\eta|$ and the same proof goes through.
EDIT: To hopefully address Ted Shifrin's comments, I'm not suggesting replacing $\eta$ with $|\eta|$ throughout, just in the RHS of the first inequality. Let me write out my proposed proof of (1.99) without any assumption on the sign of $\eta$.
By Stokes' and $\eta^2$ vanishing on $\partial \Sigma$, we know that $$0=\int_\Sigma d(2\eta^2 N^*\alpha)=\int_\Sigma 4\eta d\eta\wedge N^*\alpha+\int_\Sigma 2\eta^2dN^*\alpha.$$ We therefore compute \begin{align*} \int_\Sigma \eta^2|A|^2d\Sigma&=2\int_\Sigma\eta^2dN^*\alpha\\ &=-4\int_\Sigma\eta d\eta \wedge N^*\alpha\\ &\leq 4C\int_\Sigma |\eta| |\nabla_\Sigma \eta||A|d\Sigma\\ &\leq 4C\left(\int_\Sigma \eta^2|A|^2d\Sigma\right)^{\frac{1}{2}}\left(\int_\Sigma |\nabla_\Sigma\eta|^2d\Sigma\right)^{\frac{1}{2}}, \end{align*} and continue as CM. In particular, I don't think I'm differentiating $|\eta|$ anywhere in this proof.