An update to my earlier answer. I've written a proof of this "AC0 prime number conjecture" as a short paper, which can be found here.
https://arxiv.org/abs/1103.4991
I thought a bit about establishing a nontrivial bound on the Fourier-Walsh coefficients $\hat{\mu}(S)$ for all sets $S$. My paper does this when $|S| < cn^{1/2}/\log n$ (here $S \subseteq \{1,\dots,n\}$). On the GRH it works for $|S| = O(n/\log n)$. I remarked before that the extreme case $S = \{1,\dots,n\}$ follows from work of Mauduit and Rivat.
I still believe that there is hope of proving such a bound in general, but this does seem to be pretty tough. At the very least one has to combine the work of Mauduit and Rivat with the material in my note above, and neither of these (especially the former) is that easy.
OK, it seems that a probabilistic swapping argument works (and is simpler than the two other suggestions I made above).
Firstly, by the Bourgain-Sarnak-Ziegler criterion (Proposition 1 in http://terrytao.wordpress.com/2011/11/21/the-bourgain-sarnak-ziegler-orthogonality-criterion/ ), it suffices to show that
$$ \sum_{n \leq x} WS(pn) WS(qn) = o(x)$$
for all fixed distinct primes $p$, $q$, and in the limit $x \to \infty$.
Fix $p,q,x$. We phrase things probabilistically. Let $n$ be a random integer between $1$ and $x$; the objective is to show that
$$ {\bf E} WS(pn) WS(qn) = o(1) \quad\quad (1).$$
The way we will do this is to construct a new random variable $n'$, coupled with $n$, which asymptotically almost surely (i.e. with probability $1-o(1)$) is such that
$$ WS(pn) WS(qn) = - WS(pn') WS(qn') \quad\quad (2) $$
aas. Also, the total variation distance between the distribution of $n$ and the distribution of $n'$ will be established to be $o(1)$, so that
$$ {\bf E} WS(pn) WS(qn) = {\bf E} WS(pn') WS(qn') + o(1).$$
Putting these facts together will give the claim (1).
It remains to construct $n'$ with the desired properties. For simplicity let us begin with the case when $WS(p) = -WS(q)$. Then to obtain $n'$ from $n$, what one does is scan the binary expansion of $n$ for a block of zeroes of length at least $\ell := 10 \max( \log_2 p, \log_2 q ) + 11$ (say); note from the law of large numbers that there are going to be about $2^{-\ell} \log_2 x$ such blocks aas. We randomly select one of these blocks, and flip the middle zero in this block to a one to create $n'$. A bit of thought (using the long multiplication algorithm) shows that $WS(pn') = WS(pn) WS(p)$ and $WS(qn') = WS(qn) WS(q)$, giving (2) since $WS(p) = -WS(q)$. A rather tedious double counting involving the Chernoff inequality (which I will omit here) also shows that the distribution of n' is within o(1) in total variation to that of n, giving the claim. (The point is that the indegrees and outdegrees of the edit graph connecting potential $n$s to potential $n'$s both concentrate around the same quantity, namely $2^{-\ell} \log_2 x$.)
Of course, it could happen that $WS(p)=WS(q)$ instead. But a variant of the above argument will work as long as one can find at least one natural number $a$ for which $WS(ap) = -WS(aq)$, basically one has to insert in the binary expansion of $a$ in the middle of a sufficiently large block of zeroes of $n$ to make $n$, rather than flipping a single bit. So the only way things can go wrong is if the primes $p,q$ are such that $WS(ap)=WS(aq)$ for all natural numbers $a$ (i.e. if there is absolutely no cancellation at all in $\sum_{n \leq x} WS(pn) WS(qn)$. This can be eliminated as follows. Without loss of generality we have $p > q$ and $p$ odd. Then we can find $n,k$ such that $qn < 2^k < pn < 2^{k+1}$. We observe that
$$ WS(p (n+2^{k+1}) ) = - WS(pn) WS(p)$$
and
$$ WS(q (n+2^{k+1}) ) = WS(qn) WS(q)$$
and so we cannot have $WS(ap)=WS(aq)$ for all $a$.
Best Answer
The OP's conjecture follows from theorem 1.1 in "Primes with an average sum of digits" by Drmota, Mauduit and Rivat. They prove that the number of primes $\le x$ with $k$ binary digits is given by $$\frac{\pi(x)}{\sqrt{(\pi/2) \log x }}\left(e^{-\frac{2(k-\frac{1}{2}\log x)^2}{\log x}}+O(\log x^{-\frac{1}{2}+\epsilon})\right).$$ So inparticular this shows the stronger result in your question corresponding to the function $f(n)=\alpha \sqrt{n}$. (So one has infinitely many primes with $\frac{n}{2}+\alpha\sqrt{n}$ ones in their binary expansion.) The authors say that they weren't able to get any bounds for $f(n)=\alpha n$ with any $\alpha > 0$.