UPDATE: I notice that the question leaves the possibility of boundary open. The answer below only works as written if there is no boundary. I'm not sure what happens if there is boundary, but one should be quite careful in this case, and I would guess that the claim is false.

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The answer is "yes" if you define $\Omega^*=\exp_p(B_R(0))$ for $R$ chosen so that $|\Omega| = |\Omega^*|$. Of course, $R$ could be larger than the injectivity radius, but no worries. 

To see that you can choose such an $R$, note that $R\mapsto\exp_p(B_R(0))$ forms an exhaustion of your manifold by the complete, connected hypothesis (in fact you could probably drop compact in exchange for just requiring complete and $|\Omega| < \infty$). 


Now, to prove your identity we may use the <a href="http://en.wikipedia.org/wiki/Coarea_formula">co-area formula</a> to write (using that $|\nabla r|=1$ almost everywhere)
$$
\int_\Omega f(r) dV= \int_0^\infty f(s) Area(\{r=s\}\cap\Omega) ds
$$
and
$$
\int_{\Omega^*} f(r) dV= \int_0^\infty f(s) Area(\{r=s\}\cap\Omega^*) ds
$$
Define $\varphi(s):=Area(\{r=s\}\cap\Omega^*)-Area(\{r=s\}\cap\Omega)$. The co-area formula applied again gives that $\int_0^\infty \varphi(s) ds = 0$. Also, by construction $\varphi(s) \geq 0$ for $s\leq R$ and $\varphi(s) \leq 0$ for $s \geq R$. 

Thus, subtracting the above identities gives
$$
\int_{\Omega^*} f(r) dV-\int_\Omega f(r) dV= \int_0^\infty f(s) \varphi(s) ds
$$
$$
 = \int_0^R f(s) \varphi(s)ds + \int_R^\infty f(s)\varphi(s) ds
$$
$$
\geq f(R) \int_0^R\varphi(s)ds + f(R) \int_R^\infty \varphi(s) ds = 0.
$$

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There seems to be some confusion about the application of the co-area formula. The co-area formula holds for Lipschitz functions on a manifold. This was first proved by Federer, see Theorem 3.1 <a href="http://www.jstor.org/stable/1993504">here</a>. This reference may be difficult to follow. A more readable discussion can be found in Krantz and Parks's book "Geometric Integration Theory," see Theorem 5.4.8 and the preceding discussion.