Is there a generalization of ${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$ for arbitrary connection? I am studying general relativity and I am trying to understand how to perform variation of the Einstein–Hilbert action with respect to the metric ${{g}_{\mu \nu }}$ and an arbitrary connection ${{\omega }^{\kappa }}_{\lambda \mu }$ treating them as independent variables.
At some point I do have to perform integration by parts, where the relation $${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$$ could be useful. The problem is that I am convinced that this relation is not valid for an arbitrary connection. For the case of an arbitrary symmetric connection I have managed to prove the relation $${{\left( \sqrt{-g}{{V}^{a}} \right)}_{,\alpha }}=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}+\sqrt{-g}\frac{1}{2}{{g}^{\sigma \tau }}{{g}_{\sigma \tau }}_{;\alpha }{{V}^{\alpha }}$$ which gives $${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$$ if the connection is metric compatible. 
Therefore I would like to ask if there is a way to obtain a generalization of $${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$$ that is valid for an arbitrary connection (non-symmetric and/or not metric-compatible)?
 A: For the purposes of this answer, to adapt to the level of mathematics presented in OP, I will use (signed) densities, rather than differential forms.

Let $\rho$ be a nowhere vanishing scalar density of weight 1 (eg. of the same type as $\sqrt{-g}$, some people consider this to be weight -1 I guess). Such an object shall be called a volume density.
The covariant derivative of scalar densities of weight 1 (with respect to any connection) is defined as $$ \nabla_\mu\rho=\partial_\mu\rho -\Gamma_{\mu\nu}^\nu\rho. $$
Here $\Gamma$ doesn't denote the Christoffel symbols, but the coefficients of an arbitrary linear connection.

The connection $\nabla$ is said to be volume-presering if there is such a $\rho$ such that $\nabla_\mu\rho=0$. We make some statements regarding volume-preserving connections.


*

*If $\nabla$ is volume-preserving then $\Gamma_\mu\equiv\Gamma_{\mu\nu}^\nu$ is a gradient: $$ 0=\nabla_\mu\rho=\partial_\mu\rho-\Gamma_\mu\rho \\ \partial_\mu\rho=\Gamma_\mu\rho \\ \partial_\mu\ln\rho=\Gamma_\mu. $$ Dividing by $\rho$ was possible because we assumed it is nowhere vanishing.

*If $\nabla$ is volume-preserving and torsionless, then for any vector field $X^\mu$ we have $ \nabla_\mu X^\mu\rho=\partial_\mu(X^\mu\rho): $ $$ \nabla_\mu X^\mu=\partial_\mu X^\mu+\Gamma_{\mu\nu}^\mu X^\nu=\partial_\mu X^\mu+\Gamma_{\nu\mu}^\mu X^\nu+T^\mu_{\ \mu\nu}X^\nu=\partial_\mu X^\mu+\partial_\mu\ln\rho X^\mu+T^\mu_{\ \mu\nu}X^\nu \\ =\frac{1}{\rho}\partial_\mu(\rho X^\mu)+T^\mu_{\ \mu\nu}X^\nu. $$ Hence if the connection is torsionless, the identity is proven.

*The connection $\nabla$ is locally volume-preserving if and only if the curvature tensor is $\mathfrak{sl}(n,\mathbb R)$-valued, which in this context means that $R^\kappa_{\ \kappa\mu\nu}=0$: $$ \ $$ First we prove that if the connection is volume preserving, then the curvature is traceless: $$ R^\kappa_{\ \kappa\mu\nu}=\partial_\mu\Gamma^\kappa_{\nu\kappa}-\partial_\nu\Gamma^\kappa_{\mu\kappa}+\Gamma_{\mu\sigma}^\kappa\Gamma_{\nu\kappa}^\sigma-\Gamma_{\nu\sigma}^\kappa\Gamma_{\mu\kappa}^\sigma=\partial_\mu\partial_\nu\ln\rho-\partial_\nu\partial_\mu\ln\rho=0, $$ where the first two terms cancel because of the equality of mixed partials, and the cancellation in the second two terms is evident if the summation indices are renamed. $$ \  $$ We now illustrate that if the curvature tensor is traceless then there at least locally exists a volume density that is preserved by the connection. We have $$ 0=R^\kappa_{\ \kappa\mu\nu}=\partial_\mu\Gamma_\nu-\partial_\nu\Gamma_\mu. $$ Note that the cancellation of the last two terms doesn't depend on the specific form of the connection (it is essentially based on the identity $\text{Tr}(AB)=\text{Tr}(BA)$). By Poincaré's lemma we have then $$ \Gamma_\mu=\partial_\mu\phi $$ for some function $\phi$, at least locally. But then one may check that $$ \rho=e^\phi $$ transforms as a density, and is preserved by the connection, since $$ \nabla_\mu\rho=\partial_\mu\rho-\Gamma_\mu\rho=\partial_\mu e^\phi-\partial_\mu\phi e^\phi=0. $$ An alternative line of thought is that on densities, the curvature is $-R^\kappa_{\ \kappa\mu\nu}$, so if this vanishes, then we can at least locally (on contractible domains) integrate the partial differential equation $\nabla_\mu\rho=0$.

TLDR:
The equation $\nabla_\mu X^\mu\sqrt{-g}=\partial_\mu(X^\mu\sqrt{-g})$ does not generalize to arbitrary connection. However if the connection is such that its curvature tensor is traceless, then one may locally find a volume density $\rho$ such that if $\sqrt{-g}$ is replaced with $\rho$, and $\nabla$ is torsionless, the equation will keep holding.
