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 co-area formula to write (using that $|\nabla r|=1$) $$ \int_\Omega f(r) dV= \int_0^\infty f(s) Area(\{r=s\}\cap\Omega) ds $$ 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 $$ \int_{\Omega^*} f(r) dV-\int_\Omega f(r) dV= \int_0^\infty f(s) \varphi(s) ds $$ $$ = \int_0^R f(s) \varphi(s)ds + \int_R^\infty f(s)\varphi(s) ds $$ $$ \geq f(R) \int_0^R\varphi(s)ds + f(R) \int_R^\infty \varphi(s) ds = 0. $$