Is orientability needed to define volumes on riemannian manifolds?

First off, one of the axioms for a Riemannian metric is that the bilinear form $g_{ij}$ is positive definite and hence $\det(g_{ij})\geq 0$.

your calculation is correct -- we indeed have $$ \sqrt{\det(H(y^{-1}\circ x))}\vert J \vert =\sqrt{\det(G)}, $$ but also notice that by direct calculation, we have \begin{eqnarray*} \sqrt{\det(H(y^{-1}\circ x))} d (y^{-1}\circ x)_1 ... d (y^{-1}\circ x)_n &=& \int \sqrt{H(y^{-1}\circ x)} J dx_1 ... dx_n \\ &=& \pm \int \sqrt{H(y^{-1}\circ x)} \vert J\vert dx_1 ... dx_n \\ &=& \pm \int \sqrt{\det(G)}dx_1...dx_n\;. \end{eqnarray*} So that means that if the transformation $y^{-1}\circ x$ does not preserve orientation, we have some ambiguity in the way we define volume. On the one hand, change of variables tells us that our volumes ought to agree after transforming the metric via the transition maps. On the other hand, if we simply compute the volume after transforming the coordinates, our result differs by a sign.

You don't need orientability to define the volume of a parametrized region. Your value $vol(R)$ is completely well-defined in terms of a parametrization, and doesn't even depend on the parametrization if we take the modulus of its value, which is arguably the sensible thing to do if we are talking about "volumes". Therefore, I think that this discussion about orientability when you are restricting yourself to volumes of parametrizations can be a little uninteresting.

However, you need orientability to define a volume form on the entire manifold, so that global things like $$\int_M \omega_{vol}$$ make sense, and even so that $\omega_{vol}$ be well-defined on $M$. This is the point where orientability is important: when we talk about global phenomena. Integration in a chart is only a matter of sign choice. However, if we try to clump up the work we do in charts in order to make something in the whole manifold, these sign choices need to be chosen in such a way that things will be well-defined when we mix them up. Orientability tells us that we can do that. It is not only a matter of sign choice of the end result anymore, it is a matter of possibility, due to sign choices of building blocks, to have an end result that makes sense. If it is possible (i.e., if the manifold is orientable), then it is only a matter of sign choice of the end result.

There is already another answer dealing with the computations directly, but it may be fruitful to know that the volume form is given at each point $p$ by taking an orthogonal basis $E_1,\cdots,E_n$ with respect to $g_p$ with the correct orientation given by the orientation of the manifold, and then considering the dual basis of $T_pM^*$: $\delta_1,\cdots,\delta_n$. Then, $$(\omega_{vol})_p:=\delta_1\wedge \cdots \wedge \delta_n.$$ The fact that this is well-defined is due to the fact that changing bases will change it by a factor of a determinant. Since we defined it in terms of orthogonal basis with the same orientation, this determinant will be $1$, and thus the form is well-defined.

A straight-forward calculation shows that $\omega_{vol}$ is given by the formula you have when expressed in terms of the dual forms of an arbitrary positively-oriented chart (i.e, the $dx_i$'s).

EDIT: We can, in the non-orientable case, define $$\mu:=| \delta_1\wedge \cdots \wedge \delta_n| ,$$ with the $\delta_i$'s as before, but for an arbitrary orthogonal base $E_1,\cdots,E_n$ of $T_pM$. Since an orthogonal transformation will make a factor of $+1/-1$ appear, the modulus will kill it and it is thus well-defined as a density. Note however that $\mu$ is not a form, and by the observation of Manfredo, he has integrations of forms in mind.