Given Poisson's equation for a ball $B_r\subset\mathbb{R}^n$ \begin{align*} \Delta u &= 0 ~~~\text{in } B_r \\ u &= g~~~\text{in } \partial B_r, \end{align*} $g\in C^{\infty}$. Is it possible to show the following estimate
$$ \int_{B_r} |\nabla u|^2 dx \leq c \int_{\partial B_r} u^2dS~? $$
EDIT: the estimate does not have to hold for all $g$. I am interessed if there are $g$ so that the estimate holds. I think this was not clear.
No this is not possible. The trace of an $H^1$ function belongs to $L^p(\partial B_r)$ for some $p > 2$ (depending on the dimension $n$). Hence, your inequality would imply $$ c_1 \, \| u \|_{L^p(\partial B_r)} \le \| u \|_{H^1(B_r)} \le c_2 \, \|u\|_{L^2(\partial B_r)} $$ for some constants $c_1,c_2 > 0$ and all harmonic $u \in C^\infty(\bar B_r)$. However, this cannot be true.