I believe this is also true if $\text{deg}\ f$ equals $2$; presumably this is well-known. To prove that $f$ is unique up to isomorphism (over $\mathbb{P}^2$), it is equivalent to prove that the $\mathcal{O}_{\mathbb{P}^2}$-algebra $f_*\mathcal{O}_X$ is unique up to isomorphism. Of course, by Zariski's Main Theorem (or easier arguments), it suffices to prove uniqueness over the open complement $U$ of finitely many points of $\mathbb{P}^2$. Then there is the standard short exact sequence, $$ 0 \to \mathcal{O}_{U} \xrightarrow{f^\#} f_*\mathcal{O}_X \to \mathcal{L} \to 0,$$ where $\mathcal{L}$ is an invertible sheaf on $U$ (after deleting finitely many points). The algebra structure on $f_*\mathcal{O}_{X}$ determines an injective homomorphism of coherent sheaves, $$u:\mathcal{L}^{\otimes 2} \to \mathcal{O}_{U},$$ whose image equals the ideal sheaf $\mathcal{O}_{U}(-\underline{S})$ (again after deleting finitely many points). Thus uniqueness of $f$ is equivalent to uniqueness of the pair $(\mathcal{L},u)$ of an invertible sheaf $\mathcal{L}$ and an isomorphism $u:\mathcal{L}^{\otimes 2}\to \mathcal{O}_{U}(-\underline{S})$ up to "equivalence", i.e., $(\mathcal{L},u)$ is equivalent to $(\mathcal{M},v)$ if there exists an isomorphism of invertible sheaves $w:\mathcal{M}\to \mathcal{L}$ such that $v\circ w^{\otimes 2}$ equals $u$.

Of course for two such pairs, $(\mathcal{L},u)$ and $(\mathcal{M},v)$, then $\mathcal{T} := \mathcal{M}\otimes \mathcal{L}^\vee$ is an invertible sheaf and $v\otimes u^\dagger$ is an isomorphism $\mathcal{T}^{\otimes 2} \to \mathcal{O}_{U}$. In other words, $\mathcal{T}$ is a $2$-torsion element in $\text{Pic}(U)$. However, for every open complement $U$ of finitely many points of $\mathbb{P}^2$, $\text{Pic}(U)$ is just $\mathbb{Z}$. Thus $\mathcal{T}$ is isomorphic to $\mathcal{O}_U$, so that the pair $(\mathcal{L},u)$ is unique up to equivalence.

I am sure the case that $\text{deg}\ f$ equals $2$ is well-known. Just to make one observation: with the one exception where $X$ is a quadric surface, for $f$ as above, $f^*\mathcal{O}_{\mathbb{P}^2}(1)$ is **not** very ample. Indeed, the pullback map $$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1)) \to H^0(X,f^*\mathcal{O}_{\mathbb{P}^2}(1))$$
is an isomorphism. Of course, also $\text{deg}\ f \not\geq 12$.

positiveresults, not innegativeresults, correct? It seems to me that one can produce negative examples as a composition $f=(g_1\times g_2)\circ h$ of a degree $2$ morphism $h:\mathbb{P}^1\times \mathbb{P}^1\to \mathbb{P}^2$ with a product morphism $g_1\times g_2:C_1\times C_2 \to \mathbb{P}^1\times \mathbb{P}^1$, where $g_i:C_i\to \mathbb{P}^1$ is a simple branched cover. Did you want to add a hypothesis that $\pi_1(\mathbb{P}^2\setminus S, p) \to \text{Aut}(f^{-1}\{p\})$ is surjective? $\endgroup$ – Jason Starr Dec 23 '13 at 15:48