There is a straightforward bound.

Consider A to be the $log(N)$-fold tensor product of $H= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$.

A is Unitary and Hermitian. In fact A is just the Fourier transform.

Set $u$ to be all one vector, i.e. $u=(1, 1, 1, \ldots , 1)^T$.

And $v$ is all one vector with the first coordinate set to $-1$, i.e. $v=(-1, 1, 1, \ldots , 1)^T$

See $Au =(1, 0 , 0 , \ldots 0)^T$

and $Av$ has non zero entries in all coordinates.

Thus, $||sgn(Au)-sgn(Av)||_1 \geq N-1$.

Conclusion. No non-trivial bound.

There is a straightforward bound.

Consider A to be the $\log(N)$-fold tensor product of $H= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$.

A is Unitary and Hermitian. In fact A is just the Fourier transform.

Set $u$ to be all one vector, i.e. $u=(1, 1, 1, \ldots , 1)^T$.

And $v$ is all one vector with the first coordinate set to $-1$, i.e. $v=(-1, 1, 1, \ldots , 1)^T$

See $Au =(1, 0 , 0 , \ldots 0)^T$

and $Av$ has non zero entries in all coordinates.

Thus, $\|\operatorname{sgn}(Au)-\operatorname{sgn}(Av)\|_1 \geq N-1$.

Conclusion. No non-trivial bound.