I suppose this will follow from the plausible Cramer's conjecture about prime gap of $O(\log^2{p_n})$.
Let $n=(2^k-1)2^m$ with $m < k$ and $2^m>C\log^2{n}$ (actually $ m > \log{(C\log^2({2^m(2^k-1))}}/\log{(2)} $ will do.)
$n$ has $k$ ones and the (much) less zeros. The interval $(n,n+C\log^2{n})$ with $2^m>C\log^2{n}$ will contain a prime $p=n+\delta$. $\delta$ will contribute at least one $1$ to the zero bits of $n$ keeping all of the ones. There are infinitely many choices for $m,k$.
Not m,k$ producing distinct primes..
Legendre conjecture probably will do too.
Note that the doubly logarithmic choice of $m$ contributes relatively few zeros.
Added So if you believe Cramer's conjecture probably will do too , there are infinitely many $n$ for which $n$ bit primes have only $O(\log{n})$ zeros in their Binary expansion.
(btw, I would be very interested in unconditional answer to the question).
