I believe you get a six-to-one dominant, rational transformation from $\mathbb{P}^2$ to $X$ as soon as you have a single rational point $p$ on $X$ (except possibly if the point lies on a line or conic). Let $C'$ be the intersection of $X$ with the tangent $2$-plane $\Lambda_p$ to $X$ at $p$. The normalization $C$ of $C'$ is a genus $0$ curve that admits a degree $3$ zero-cycle, namely the inverse image of a hyperplane section of $C'$. Thus $C$ is a rational curve.

Consider the $\mathbb{P}^2$-subbundle $\Lambda \subset C \times \mathbb{P}^3$ of tangent $2$-planes $\Lambda_q$ to $X$ at points $q$ of $C$. The intersection of $\Lambda$ with $C\times X$ is a non-normal surface $S'$ fibered over $C$. The normalization $S$ is a conic bundle over $C$ that admits a degree $3$ zero-cycle, namely the inverse image of a hyperplane section. Thus $S$ is a rational conic bundle over the function field of $C$. In particular, $S$ is rational.

By my computations, the degree of the projection morphism from $S$ to $X$ equals $6$.