OK, let $A$ be a subset of *even* vertices of cardinatily $2^n\mu\ge 2^{n-2}$ (so $\mu\ge \frac 14$) and let $n\ge 3$. Let $B$ be the set of odd vertices with at least 2 neighbors in $A$. Assume that $|B|=2^n\xi$. Our aim is to show that $\xi\ge\mu$. Let us consider the action of the averaging (over neighbors) operator $T$ in $L^2$ with respect to the Haar measure. 

Let $f$ be the characteristic function of $A$. Let $g$ be $f$ with the constant and the alternating components removed. Then $\|g\|^2=\mu-2\mu^2$ and, thereby, $\|Tg\|^2\le (1-\frac 2n)^2(\mu-2\mu^2)$ because we removed the eigenspaces corresponding to the eigenvalues $\pm 1$ and every other eigenvalue is at most $1-\frac 2n$.

On the other hand, we know that $Tg\le \frac 1n-2\mu$ on the complement of $B$ in the set of odd vertices. To balance it to the average $0$, we should have $Tg$ at least $\frac{\frac 12-\xi}{\xi}(2\mu-\frac 1n)$ on $B$ on average and, since the quadratic average can be only larger, we get
$$
\|Tg\|^2\ge \left[(\frac 12-\xi)+\xi\left(\frac{\frac 12-\xi}{\xi}\right)^2\right]
(2\mu-\frac 1n)^2=\frac 12\frac{\frac 12-\xi}{\xi}(2\mu-\frac 1n)^2
$$
Thus
$$
\frac 12(\frac 12-\xi)(2\mu-\frac 1n)^2\le \xi(1-\frac 2n)^2(\mu-2\mu^2)
$$
Now, $2\mu-\frac 1n=2\mu(1-\frac{2}{4\mu n})\ge 2\mu(1-\frac 2n)$ under our assumption $\mu\ge \frac 14$. So, we get
$$
\frac 12(\frac 12-\xi)4\mu^2 \le \xi(\mu-2\mu^2)
$$
or, equivalently,
$$
(1-2\xi)\mu\le (1-2\mu)\xi
$$
i.e.,
$$
\mu\le\xi.
$$

I hope I haven't made a stupid mistake anywhere though I do not really like this proof: it works for $\stackrel{2}+$, but not for $\stackrel{4}+$ and you are, probably, interested in $\stackrel{K}+$ for all fixed $K$ as $n\to\infty$. Anyway, it gives the desired cutoff at $1/2$ for fixed parity and, thereby, the cutoff at $\frac 34$ in general.