$A=[0,1]$ has positive measure and is not locally $\varepsilon$-dense for $(x,y)=(3,4)$ ...
For the converse, yes and even more $\mathbf{R}\setminus A$ has measure $0$. And you only need the estimate for one single $\varepsilon$ :
$\mathbf{1}_A$ is locally integrable si almost every $x\in \mathbf{R}$ is a Lebesgue point, that is
\begin{align*}
\lim_{r\rightarrow 0^+} \frac{1}{2r}\int_{[x-r,x+r]} \mathbf{1}_A = \mathbf{1}_A(x).
\end{align*}
The local $\varepsilon$-density imposes that $\mathbf{1}_A>0$ almost everywhere.