Surely Gauss knew about this. Consider $S \subset \mathbb R^3$, a two dimensional oriented embedded surface. Recall that the Gauss map $G: S \to S^2$ is defined by $G(p) = \mathbb n(p)$ where $\mathbb n(p)$ is the unit normal of $S$ at $p$, translated to the origin (so that $\mathbb n \in S^2$, the unit sphere). Then the Gaussian curvature at $p$ effectively measures the signed area distortion of a small rectangle near $p$ under the map $G$. Thus the Gauss curvature $K$ is defined to be $\textrm{det}(DG)$ where $DG$ is the derivative of $G$, say, written with respect to a local parametrization of $S$ in some neighborhood of $p$. The hypothesis that $K := \textrm{det}(DG) \geq 0$ is thus saying something about area deformation. So it is not a long walk from here to the proof that you want (most proofs also involve one other concept: normal (or "geodesic") coordinates, also invented by Gauss). As far as a reference is concerned, consider looking at Ted Shifrin's great book Differential Geometry. I don't know if he proved the volume comparison result you wanted per se, but the "pieces" are certainly there.
Edit: After a little bit more thought, I guess it is more difficult than I thought to outright prove this from first principles based on what I said above. I don't know whether or not I've ever seen it worked out completely just for, say, surfaces in $\mathbb R^3$ (the more general results imply this case of course, but that is not what your question is about).