Let me prove that for $f\in L^\infty$ and fixed $\delta>0$ (and centered balls) such $C$ exists. The idea is to find $C$ which works for all Lebesgue points of $f$. Assume that there is no such $C$, then for $C=n$ there exist violating Lebesgue points $x_n$ and $B_n$ with center $x_n$ and radius $r_n\geqslant \delta$ and $$|f(x_n)|> n\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B_n}|f|.\tag{$\heartsuit$}$$ Without loss of generality, $x_n$ converge to a point $x_0$ (it lies in $\Omega$, since all $x_n$ are at distance at least $\delta$ from $\mathbb{R}^n\setminus \Omega$, thus so does $x_0$). Then for the ball $B(x_0,\delta)$ we have $\int_{B(x_0,\delta)}f=0$: otherwise the integrals over $B_n$ would be bounded from below, and RHS's of $(\heartsuit)$ would tend to infinity, contradicting to LHS's being bounded from above by $\|f\|_\infty$. But for large $n$ we have $x_n\in B(x_0,\delta)$, and, since by $(\heartsuit)$ we have $f(x_n)\ne 0$ and $x_n$ is a Lebesgue point of $f$, we conclude that $f$ is non-zero on some set of positive measure inside $B(x_0,\delta)$. A contradiction.
For other direction, for centered balls this trivially fails for example for $f(x)=x^{-1/2}$, $\Omega=(0,1)$ (for $x$ near the singularity there are simply no centered balls of radius $\geqslant \delta$ contained in $\Omega$). For not centered balls it is However true, and settled in comments, the key word is uniform interior sphere.