The usual physical interpretation of the Reimann tensor $R^\mu_{\ \ \nu \rho \sigma}$ at a given point $p$ on a manifold $M$ is that it inputs an infinitesimal vector $v^\nu$ and two other infinitesimal vectors $x^\rho$ and $x^\sigma$, the latter of which determine an infinitesimal parallelogram-shaped loop in $M$ anchored at $p$, and outputs the amount by which $v^\nu$ changes after being parallel-transported around the infinitesimal loop. That is, I believe it gives the holonomy of infinitesimal oriented loops with base point $p$ lying in the $x^\rho$-$x^\sigma$ plane.
There are other, equivalent ways to think about what type of mathematical object the Riemann curvature tensor is. E.g. one way to think of it as a self-adjoint operator on the real inner product space of two-forms. (This interpretation is arguably more natural for the "all-lowered" version $R_{\mu \nu \rho \sigma}$.)
Is it also possible to think of the Riemann tensor as a linear-operator-valued two-form in its last two indices $\rho$ and $\sigma$, and therefore integrate it over a 2D surface in the manifold? I've only ever seen the Riemann tensor described in terms of infinitesimal loops in the manifold. But is it legitimate to think of it as a two-form and integrate it (over $\rho$ and $\sigma$) over a finite surface $S$, and get the holonomy of the oriented boundary loop $\partial S$?
On the one hand, this seems plausible by the general heuristic picture for Green's theorem:
On the other hand, I doubt that it's correct, because (a) I can't find any sources saying that it is, and (b) more concretely, it seems to me that the holonomy of an oriented loop $\partial S$ should depend on the base point (although I'm not sure about that). But the formal integral $$ \iint_S R^\mu_{\ \ \nu \rho \sigma} dx^\rho \wedge dx^\sigma $$ does not pick out any distinguished point on $\partial S$.
Is there any sense in which this is a legitimate operation? If not, is there any other way to construct the (pseudo-)Riemannian holonomy of arbitrary oriented loops (with base point) in $M$ from the Riemann tensor? Or does that require knowing the connection and/or the full metric tensor?
I know very little about tensor-valued differential forms, so this might all be complete nonsense. I suspect the fact that the Riemann tensor is an actual tensor means that this doesn't work, because for tensor-valued forms the "form" indices and the "tensor" indices somehow transform differently under coordinate transformations.
Edit: Apparently the Riemann tensor can be thought of as a Lie-algebra-valued two-form (that takes values in $\mathfrak{so}(n)$) called the curvature form. I don't know how integration of Lie-algebra-valued differential forms works, though.