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
$$
\int_{\Omega^*} f(r) dV-\int_\Omega f(r) dV= \int_0^\infty f(s) \varphi(s) ds
$$
$$
 = \int_0^R f(s) \varphi(s)ds + \int_R^\infty f(s)\varphi(s) ds
$$
$$
\geq f(R) \int_0^R\varphi(s)ds + f(R) \int_R^\infty \varphi(s) ds = 0.
$$

---


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.