The answer is no. Let us consider the case where $f$ is birational. In this case, for each point $x\in \mathbb{P}^2$ such that $f$ is not defined, the intersection of $\Gamma$ with $x\times \mathbb{P}^2$ contains all curves of $\mathbb{P}^2$ that are contracted by the inverse $f^{-1}$. The degree of these curves can be arbitrary large.

As an example, you can consider some so-called "Halphen map", that is a birational map of $\mathbb{P}^2$ that preserves a pencil of curves of genus $1$ (a simple case is given by a pencil given by two general cubics). You blow-up the $9$ points $p_1,\ldots,p_9$ of intersection and consider a translation given by $E_1-E_2$, where $E_1$ and $E_2$ are the exceptional divisors of $p_1$ and $p_2$ (which are two sections of the pencil). If you consider this map $f$ on $\mathbb{P}^2$, the degree of $f^n$ grows quadratically with $n$ and there are exactly $9$ points not defined for $f^n$, namely $p_1,\ldots,p_n$. There are exactly $9$ curves contracted by $f^{-n}$ and these are irreducible of degree that is also growing. For $n$ large, the degree of the curves contracted is then unbounded.