First, we will find the domain of $f(x)$. From $x^2 - 2x - 3 > 0$, we obtain $x < -1$ or $x > 3$. So the domain of definition for $f(x)$ is $(-\infty, -1) \cup (3, +\infty)$.

But $u = x^2 - 2x - 3 = (x-1)^2 - 4$ is monotone decreasing on $(-\infty, -1)$, and monotone increasing on $(3, +\infty)$.

So $f(x) = \log_{\frac{1}{2}}(x^2 - 2x - 3)$ is monotone increasing on $(-\infty, -1)$, and monotone decreasing on $(3, +\infty)$.

Answer: A.