(I suspect that this question has an elementary resolution. But perhaps it would be more appropriate on MathOverflow, and if so I would not be opposed to migrating it there.)
Let $V \Subset U$ be domains in a Riemannian manifold $M$. If $u: U \to \mathbb R$ is a harmonic function, then we have Caccioppoli's inequality $$\int_V |du|^2 \lesssim \int_{U \setminus V} |u|^2.$$ Possibly after adding a constant term to $u$, and applying Poincaré's inequality, $$\int_V |du|^2 \lesssim \int_{U \setminus V} |du|^2.$$ Now let $F$ be a real-valued $2$-form on $U$, which satisfies the Yang-Mills (or Maxwell) equations $dF = 0$ and $d^* F = 0$, where $d^*$ is the codifferential on $M$. If we choose a gauge field $A$, thus $F = dA$, and reason as in the proof of Caccioppoli's inequality, we obtain $$\int_V |F|^2 \lesssim \int_{U \setminus V} |A|^2.$$ Unfortunately this inequality is not very useful because it is not gauge-invariant.
I would love to be choose a gauge such that $$\int_{U \setminus V} |A|^2 \lesssim \int_{U \setminus V} |F|^2,$$ and conclude the titular "gauge-invariant Caccioppoli inequality" $$\int_V |F|^2 \lesssim \int_{U \setminus V} |F|^2.$$ Does this inequality hold?
Here's some preliminary thoughts. The natural choice of gauge is Coulomb gauge $d^* A = 0$, in which the Yang-Mills equation reduces to the Laplacian $\Delta A = 0$. We still have the gauge freedom to replace $A$ by $B := A + d\varphi$ for a harmonic function $\varphi$. My intuition is that by choosing $\varphi$ so that $B$ has "zero mean on $U \setminus V$" (yes, I know that the curvature of $M$ implies that this is rather dodgy) we obtain $$\int_V |F|^2 \lesssim \int_{U \setminus V} |B|^2 \lesssim \int_{U \setminus V} |\nabla B|^2.$$ Since $B$ solves the Dirac equation $(d + d^*) B = dA$, which is elliptic, we should have something like $$\int_{U \setminus V} |\nabla B|^2 \lesssim \int_{U \setminus V} |F|^2 + \int_{\partial U} |B|^2 + \int_{\partial V} |B|^2,$$ which is almost what we want. But I seem to be upriver without a paddle: there are boundary terms and it's not clear what we can do with them.