$\newcommand\Om\Omega\newcommand\th\theta\newcommand\p\partial$After the previous answer was given, the OP changed "Lipschitz connected" to "uniform Lipschitz". If the latter is understood in the sense of the Wikipedia definition with $r$, $h$, and the Lipschitz constant (say $K$) the same for all points $p\in\p\Om$, then the answer becomes yes.
Indeed, then for each $p\in\p\Om$ there is a unit vector $u$ such that for all $s\in(0,r]$ $$|B_p(s)\cap\Om|\le|B_p(s)\cap C_{p,u,K}|=s^d|B_p(1)\cap C_{p,u,K}| =s^d(1-\th)|B_p(1)|=(1-\th)|B_p(s)|, $$ where $d$ is the dimension of the ambient Euclidean space (say $E$), $|\cdot|$ is the Lebesgue measure, $B_p(s)$ is the open ball in $E$ centered at $p$ of radius $s$, $$ C_{p,u,K}:=\{x\in E\colon u\cdot(x-p)\le K\|\pi_{u^\perp}(x-p)\|\},$$ $\cdot$ is the inner product, $\pi_{u^\perp}$ is the orthoprojector onto the orthogonal complement to $u$, $\|\cdot\|$ is the Euclidean norm (so that $C_{p,u,K}$ is a cone, with vertex $p$), and $\th:=1-|B_p(1)\cap C_{p,u,K}|/|B_p(1)|$, so that $\th$ is a strictly positive real number, which depends only on $K$ (but not on $p$ or $u$).
Thus, we do have $$|B_p(s)\cap\Om|\le(1-\th)|B_p(s)|$$ for all $p\in\p\Om$ and all $s\in(0,r]$.