Let $(M,g)$ be a non-compact Riemannian manifold of finite volume and dimension $d$, i.e. $\mathrm{vol}_{g}(M)<\infty$. Is it always possible to find a conformal transformation $g\mapsto \widetilde{g}:=\Omega^{2}g$ for some $\Omega\in C^{\infty}(M)$ such that $\mathrm{vol}_{\widetilde{g}}(M)=\infty$?
Since $\mathrm{det}(\widetilde{g})=\Omega^{2d}\mathrm{det}(g)$, the volume of $M$ with respect to $\widetilde{g}$ is given by $$\mathrm{vol}_{\widetilde{g}}(M)=\int_{M}\vert\Omega\vert^{d}\,\mathrm{vol}_{g},$$
so the question is probably equivalent to ask whether there exist non-trivial functions $\Omega\in C^{\infty}(M)$ such that $\Omega\notin L^{d}(M,g)$.
Yes, this is definitely possible. The idea is just to make a bunch of super tall spikes. Perhaps there are cleaner ways of “implementing” such spikes, but here’s one way.
Since $M$ is non-compact, there is an open cover $\{U_{\alpha}\}_{\alpha\in A}$ with no finite subcover. Let $\{\phi_{\alpha}\}_{\alpha\in A}$ be a smooth partition of unity subordinate to this open cover. Let $\{c_{\alpha}\}_{\alpha\in A}$ be a collection of positive numbers, to be decided soon. Now, consider the function \begin{align} f&:=1+\sum_{\alpha\in A}c_{\alpha}\phi_{\alpha}. \end{align} One of the properties of a partition of unity is that the supports $\{\text{supp}(\phi_{\alpha})\}_{\alpha\in A}$ are locally finite. This implies that the sum converges and the result is a smooth function which is strictly positive (since I added the $+1$).
Let $\mu$ be any positive measure on $M$ which assigns positive mass to non-empty open sets (for example the Riemannian volume measure). Then, by a simple monotonicity argument, \begin{align} \int_Mf\,d\mu&\geq \sum_{\alpha\in A}c_{\alpha}\int_M\phi_{\alpha}\,d\mu.\tag{$*$} \end{align} Now I claim that there are infinitely many $\phi_{\alpha}$’s which are not identically zero. To see this, suppose for contradiction only finitely many of them are not identically zero, say $\phi_{\alpha_1},\dots,\phi_{\alpha_k}$. Then, the corresponding open sets $U_{\alpha_1},\dots, U_{\alpha_k}$ must cover $M$ because for each $x\in M$, we have \begin{align} 1=\sum_{\alpha}\phi_{\alpha}(x)=\sum_{i=1}^k\phi_{\alpha_i}(x), \end{align} so by non-negativity of the $\phi$’s atleast one of the terms on the right is positive, which implies that $x\in\text{supp}(\phi_{\alpha_i})\subset U_{\alpha_i}$ for some $i$, i.e we have shown that $M$ is covered by $U_{\alpha_1},\dots, U_{\alpha_k}$. Hence, we’ve arrived at the desired contradiction.
Now, discard those indices in $A$ for which the $\phi_{\alpha}$ vanishes identically (the function $f$ is clearly unchanged by this notational abuse). Hence, we may now assume that we have an infinite index set $A$ and that for each $\alpha\in A$, the function $\phi_{\alpha}$ is not identically zero. By continuity, there is thus a non-empty open set on which $\phi_{\alpha}$ is positive, i.e there is a set of positive $\mu$-measure on which $\phi_{\alpha}$ is positive, so $\int_M\phi_{\alpha}\,d\mu>0$. So, in $(*)$ we have infinitely many positive terms, so by taking $c_{\alpha}$ to be large enough (actually $c_{\alpha}=\frac{1}{\int_M\phi_{\alpha}\,d\mu}$ works) we can ensure that the RHS of $(*)$ diverges, and hence $\int_Mf\,d\mu=\infty$. Therefore, $f\notin L^1(\mu)$. Finally, by taking $\Omega=f^{1/p}$ (where $1\leq p<\infty$) we have that $\Omega$ is a smooth positive function on $M$ such that $\Omega\notin L^p(\mu)$.