$\newcommand{\tb}{\tilde B}$
Let $d:=n$.
The dispersion condition
\begin{equation*}
\lim_{r\downarrow0}\frac{|E\cap B_r(x)|}{|B_r(x)|}=0
\end{equation*}
is of no help, where $|\cdot|$ denotes the Lebesgue measure on $\mathbb R^d$.
More specifically, the following is true:

**Theorem** Suppose that $f$ is a nonnegative function in $L^1(B_1)$ such that
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{B_r}f}{|B_r|}=\infty, \tag{0}
\end{equation*}
where $B_r:=B_r(0)$, the open ball of radius $r$ centered at $0$. Then one can construct a measurable set $E\subset B_1$ such that
\begin{equation*}
\lim_{r\downarrow0}\frac{|E\cap B_r|}{|B_r|}=0 \tag{1}
\end{equation*}
but
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{E\cap B_r}f}{|B_r|}=\infty. \tag{2}
\end{equation*}

So, without any conditions on $E$ in addition to the dispersion condition (1), the best sufficient condition for
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{E\cap B_r}f}{|B_r|}<\infty \tag{not-2}
\end{equation*}
is the trivial sufficient condition
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{B_r}f}{|B_r|}<\infty. \tag{not-0}
\end{equation*}

To simplify the presentation of the proof of this theorem a bit, assume that $d=2$. By (0), there is a sequence $(r_n)$ decreasing to $0$ such that
\begin{equation*}
\int_{B_{r_n}}f\ge 2^n|B_{r_n}|
\end{equation*}
for all natural $n$. So, passing successively to subsequences, we can construct an increasing sequence $(n_k)$ of natural numbers and a sequence $(S_k)$ of sets such that
\begin{equation*}
n_k\ge2k,
\end{equation*}
\begin{equation*}
\int_{S_k}f\ge 2^{-k}2^{n_k}|\tb_k|\ge2\int_{\tb_{k+1}}f
\end{equation*}
with
\begin{equation*}
\tb_k:=B_{r_{n_k}},
\end{equation*}
and, for each natural $k$, $S_k$ is a sector of the disk $\tb_k$ with the central angle $2\pi/2^k$ such that $S_k\supset S_{k+1}$.
Let now
\begin{equation*}
E:=\bigcup_k(S_k\cap(\tb_k\setminus\tb_{k+1}))
=\bigcup_k(S_k\setminus\tb_{k+1}).
\end{equation*}
Then for any natural $k$ the condition $r_{n_{k+1}}\le r\le r_{n_k}$ implies
$E\cap B_r\subseteq S_k\cap B_r$, so that $|E\cap B_r|\le|S_k\cap B_r|=2^{-k}|B_r|$, which shows that (1) holds.

On the other hand,
$$E\cap\tb_k\supseteq E\cap(\tb_k\setminus\tb_{k+1})=S_k\cap(\tb_k\setminus\tb_{k+1})=S_k\setminus\tb_{k+1},$$ whence
\begin{multline*}
\int_{E\cap\tb_k}f
\ge\int_{S_k\setminus\tb_{k+1}}f
\ge\int_{S_k}f-\int_{\tb_{k+1}}f \\
\ge\frac12\int_{S_k}f
\ge2^{-k-1}2^{n_k}|\tb_k|\ge2^{k-1}|\tb_k|.
\end{multline*}
So, (2) also holds. $\Box$