My guess is that this is true, however there is basically no hope in proving this at all in general. The most relevant conjecture here is Manin's conjecture [1].

As an example, take $S$ to be a smooth surface of general type. Then Lang's conjecture predicts that the rational points on $S$ are not Zariski dense. In particular, the closure $Z \subset X$ of the set of rational points consists of a finite union of curves on $S$. The number of rational points on $S$ of height at most $T$ which do not lie on a rational curve is therefore $O( (\log T)^{r/2})$, where $r$ is the maximum of the Mordell-Weil ranks of the elliptic curves contained in $Z$. In particular what you want should be true in this case.

However Lang's conjecture is only known in some very special cases. E.g. it is unknown for smooth surfaces of degree $d \geq 5$ in $\mathbb{P}^3$. For such surfaces one has no idea how to prove sub-polynomial growth away from the rational curves. The best known method for getting upper bounds here is the determinant method of Bombieri-Pila/Heath-Brown/Salberger, however this gives good results for the "worst case", so it does not know the difference between a smooth surface and a singular one. (It can be used to throw away rational curves, but will still give polynomial bounds for the remainder).

The case of K3 surfaces (e.g. smooth quartic surfaces in $\mathbb{P}^3$), is very interesting as here the rational points are expected to be Zariski dense as soon as there is a single rational point. Here Manin's conjecture [1,Thm. 3.5] predicts that for all $\delta > 0$, there exists an open dense subset $U_\delta \subset S$ such that $U_\delta$ contains $O(T^\delta)$ rational points of height at most $T$. Note that in the conjecture as $\delta$ decreases it says that you need to throw away more and more curves (as $U_\delta$ depends on $\delta$). However if $S$ contains no rational curves, then you see that at each step you are only removing finitely many curves of positive genus. From this one easily sees that the result you want follows from Manin's conjecture in this case. However again no one can prove this conjecture for any K3 surface!

[1] - Batyrev, Manin - Sur le nombre des points rationnels de hauteur borné des variétés algébriques

3more comments