No. Only the upper bound holds, and as Dirk pointed out the best constant is $C_1=1$. The lower bound cannot hold for any $C_0$, since otherwise the $L^p$ and $L^q$ norms would be equivalent. This is well-known to fail. To see this, observe that with your assumptions your reference measure $\pi(x)dx$ is locally equivalent to the Lebesgue measure $dx$ (i-e $c_0 dx\leq \pi(x)dx\leq c_1 dx$ on any compact set $J\subset\mathbb R^d$). The comparison between $L^p$ and $L^q$ norms on $\mathbb R^d$ are well known for the Lebesgue measure: Taking a radial function $g(x)=\chi_{B_R}(x) |x|^\alpha$ and tuning $\alpha$ (denpending on $p,q$) easily gives a counterexample to $\|g\|_{L^p}\leq C_1\|g\|_{L^q}$, i-e such that $\|g\|_{L^p}=+\infty$ but $\|g\|_{L^q}<\infty$. (Here I'm using a cutoff function to localize on a ball $B_R$ so that $\pi(x)dx$ and $dx$ are equivalent).

No. Only the upper bound holds, and as Dirk pointed out the best constant is $C_1=1$. The lower bound cannot hold for any $C_0$, since otherwise the $L^p$ and $L^q$ norms would be equivalent. This is well-known to fail. To see this, observe that with your assumptions your reference measure $\pi(x)dx$ is locally equivalent to the Lebesgue measure $dx$ (i-e $c_0 dx\leq \pi(x)dx\leq c_1 dx$ on any ball $B_R$). The comparison between $L^p$ and $L^q$ norms on $\mathbb R^d$ is well known for the Lebesgue measure: Taking a radial function $g(x)=\chi_{B_R}(x) |x|^\alpha$ and tuning $\alpha<0$ (denpending on $p,q$) easily gives a counterexample to $C_0\|g\|_{L^p}\leq \|g\|_{L^q}$, i-e such that $\|g\|_{L^p}=+\infty$ but $\|g\|_{L^q}<\infty$. (Here I'm using a cutoff function to localize on a ball $B_R$ so that $\pi(x)dx$ and $dx$ are equivalent). More explicitly, take $\alpha=-\frac{d}{q}+\epsilon$ (for very small $\epsilon>0$) so that $|g(x)|^qdx\sim r^{\alpha q}r^{d-1}dr=r^{\alpha q+d-1}=r^{-1+\epsilon}dr$ is bordeline integrable at the origin, while $|g(x)|^pdx\sim r^{\alpha p+d-1}dr$ is not (due to the exponent $\alpha p+d-1<-1$ since $p>q$).