A version of Gauss's theorem says that
If $U$ is a bounded, open subset of $\mathbb{R}^n$ with $C^1$ smooth $\partial U$, then for $u \in C^1 (\bar U)$ $$ \int_U u_{x_i} \, dx = \int_{\partial U} u v^i \, dS $$ where $v^i$ denotes the $i$th component of outward normal to $\partial U$.
(1) Are these integrals Lebesgue integrals, or are they Riemann integrals?
(2) If the integrals can be interpreted in both Lebesgue and Riemann sense, are they the same?
(3) In Lebesgue sense, $dx$ is the Lebesgue meausure in $\mathbb{R}^n$ but which Lebesgue measure is used for the "surface element $dS$?
Note that the conditions require that $u$ is continuous and $\partial U$ is $C^1$ smooth (i.e., it can be represented locally by continuous maps). Under those conditions, Riemann integration is well-defined. And anywhere Riemann integration is well-defined, so is Lebesgue integration and the two are equal in value.
So if you want to consider it Lebesgue integration, you are free to do so. If you want to consider it Riemann integration, you are also free to do so. If you prefer a different integration, it will almost certainly work as well, as any worthwhile itegration theory is going to be able to handle continuous functions.
The surface element $dS$ is the area measure/differential on the surface.