Okay, so basically the answer can be found in here: http://arxiv.org/pdf/math/0512608v1.pdf

Here's how the argument works (a simplified version of what is done in the paper with finite dimensions and $p=q$):

*(Note: we define "$\Re$" of a vector by taking the real part componentswise)*

Lemma 3.4 says (applied to the finite dimensional situation)
$$
\int_0^{2\pi} \| \Re(e^{i\varphi} x) \|_p^p d\varphi = \int_0^{2\pi} |\cos(\varphi)|^p d\varphi
$$
for any $x\in \mathbb C^n$ with $\|x\|_p=1$. This is fairly elementary to verify.

Therefore, whenever $x,y\in \mathbb C^n$ both have norm $1$, we will find a $\varphi\in[0,2\pi]$ such that $$\|\Re(e^{i\varphi}x)\|_p \leq \|\Re(e^{i\varphi}y)\|_p$$
since the integral
$$
\int_0^{2\pi}\|\Re(e^{i\varphi}y)\|_p^p - \|\Re(e^{i\varphi}x)\|_p^p d\varphi = 0
$$
is zero und thus the integrand has to be non-negative somewhere.

Then Lemma 3.2 of the paper yields the result. What the authors do here is to take a vector
$0\neq x\in \mathbb C^n$ such that $\|Ax\|_p/\|x\|_p$ is maximal *(assume also that $A\neq 0$; then $Ax\neq 0$ follows automatically and we can divide by its norm below)*. Then they take a $\varphi$ such that
$$
\left\|\Re(e^{i\varphi} \frac{x}{\|x\|_p})\right\|_p \leq \left\| \Re(e^{i\varphi} \frac{Ax}{\|Ax\|_p})\right\|_p
$$
which is possible by the above. If $\Re(e^{i\varphi}x)\neq 0$, this can then be rewritten as
$$
\frac{\|Ax\|_p}{\|x\|_p} \leq \frac{\|A \Re(e^{i\varphi}x)\|_p}{\|\Re(e^{i\varphi}x)\|_p}
$$
which shows that the maximum is also attained at the real vector $\Re(e^{i\varphi}x)\in \mathbb R^n$.
If $\Re(e^{i\varphi}x)=0$, then $i\cdot e^{i\varphi}x$ is a real vector at which the maximum is attained.