Consider the definition of the Gamma function
$$ \Gamma(s) = \int_{0}^{\infty}\left[x^{s-1}e^{-x} \right] dx $$
Clearly: $x^{s-1}$ may have multiple defined values for $s$ if $s-1$ is rational or even infinitely many if $s-1$ is irrational.
Does that mean that when we look at the Gamma function we are really only observing a specific branch cut of it?
No, $\Gamma$ has no branch cuts. That definition can only be taken for $s$ whose real part is positive, as the integral diverges otherwise. Noting that we define $x^s=e^{s\log(x)}$, we are tempted to worry about the fact that $\log$ is multi-valued - but this proves to be a non-issue. Why? Because the principal value of $\log$ is continuous in the left half-plane - there are no branch cuts there. Given that the gamma function, from the left half plane, can be continued (analytically) to the whole plane using the identity $\Gamma(t+1)=t\Gamma(t)$, which introduces no branch cuts, it's clear that the function as a whole must have no branch cuts.
You might be tempted to worry about what happens if we take $x^s=e^{s(\log(x)+2n i \pi)}$ for some integer $n$ - and the answer is that we get another meromorphic function, also satisfying $f(t+1)=tf(t)$ and $f(1)=1$. It would, in fact, satisfy $f(s)=e^{(s-1)2n i \pi}\Gamma(s)$ - but it is a distinct function.