Schwarz's lemma is somewhat equivalent to having $d^2=0$, where $d$ is the exterior derivative on a $C^k$ manifold, $1 \leq k \leq \infty$. However I can't see how this could fail, for, say, $f$ a twice-differentiable but not necessarily $C^2$ function. After all this is just saying that the boundary of a boundary is empty, and I don't see how the irregularities of $f$ could result in a non-zero integral.
To be more precise, let's say $f$ is a twice-differentiable (i.e. $\mathrm{d}f$ exists and is differentiable) function of two real variables. I'd like to know if it's true that $$\int_U \mathrm{d}^2f=0$$ for all open subsets of the plane $U$. For this it would suffice to have the Schwarz lemma be true of $f$ almost everywhere (in the sense of Lebesgue), and incidentally all the counterexamples I know of do fit this description.
For a function $f:U \subset \Bbb{R}^n \to \Bbb{R}$, you don't need $C^2$ for Schwarz's lemma (i.e equality of mixed partials). All you need is for $f$ to be twice Frechet differentiable, which means $Df:U \to \mathcal{L}(\Bbb{R}^n, \Bbb{R})$ exists and is itself differentiable. In fact, you can formulate the lemma pointwise, so if the second derivative exists at at one point then the mixed partials at that point are equal.
More generally, if you have two real Banach spaces $E_1, E_2$, an open subset $U\subset E_1$, and a twice-differentiable map $f:U \to E_2$, then the second (Frechet) derivative at a point $a \in U$, $D^2f_a$ can be thought of as a bilinear map $E_1 \times E_1 \to E_2$. One can prove that this bilinear map is symmetric (which in the finite-dimensional case is equivalent to the equality of mixed partial derivatives).
For a proof of these statements, see Henri Cartan's Differential Calculus text, Sections $5.1, 5.2$.