Let $B_r$ be an open ball in $\Bbb R^d$ with radius $r$. Let $u\in W^{1,p}(B_r)$ and define $ \bar u=\frac 1{|B_r|}\int u\ dx, $ so that by Poincare inequality we have $$ ||u-\bar u||_{L^p(B_r)} \le C||Du ||_{L^p(B_r)}\quad;\quad C=C(r,p). $$ Actually, $C(r,p)=C(p)\cdot r$ by dilating $B_r$ to $B_1$ and change of variables.
Now, by Sobolev embedding we have $W^{1,p}(\Omega) \hookrightarrow L^q(\Omega)$ for $q\in [1,\frac{dp}{d-p}]$, i.e. $$ ||u||_{L^q} \le C'||u||_{W^{1,p}}\quad;\quad C'=C'(\Omega,p,q,d). $$
Combining these we can easily get $$ ||u-\bar u||_{L^q(B_r)} \le C'' ||Du||_{L^p(B_r)} $$ for some constant $C''$ depending on $r,p,q,d$.
However, I've seen a version that says $$ \left(\frac 1{|B_r|}\int |u-\bar u|^q \right)^{1/q} \le Cr \left(\frac 1{|B_r|}\int |Du|^p \right)^{1/p} $$ where $C=C(d,p,q)$ does not depend on $r$.
Is that an easy consequence of the above or does it require another approach entirely? I tried scaling argument but the term $||u||_p$ and $||Du||_p$ in $||u||_{W^{1,p}}$ scale differently. I seem to somewhat stuck, could anyone please help?
The inequality you want is only true for $q=\frac{dp}{d-p}$. Assume first that $g\in W^{1,p}(B_1)$. Then you can extend $g$ to a function $G\in W^{1,p}(\mathbb{R}^d)$ with $$\Vert G\Vert_{W^{1,p}(\mathbb{R}^d)}\le c \Vert g\Vert_{W^{1,p}(B_1)}.$$ Next apply Sobolev embedding to find $$\Vert g\Vert_{L^{p^*}(B_1)}\le\Vert G\Vert_{L^{p^*}(\mathbb{R}^d)}\le c \Vert \nabla G\Vert_{L^{p}(\mathbb{R}^d)}\le C(\Vert \nabla g\Vert_{L^{p}(B_1)}+\Vert g\Vert_{L^{p}(B_1)}).$$ Now you apply this inequality to $u-\bar u$ to get $$\Vert u-\bar u\Vert_{L^{p^*}(B_1)}\le C(\Vert \nabla u\Vert_{L^{p}(B_1)}+\Vert u-\bar u\Vert_{L^{p}(B_1)}) \le C(\Vert \nabla u\Vert_{L^{p}(B_1)},$$ where in the last inequality you apply Poincare's inequality (the one you wrote at the beginning).
Now you do a rescaling and use the fact that $p^*=\frac{dp}{d-p}$.