This depends (again) on $n$ and $s$. Consider the case where $n=2$. If $s<9$ then $-K_X$ is ample and the answer is yes. If $s=9$ then you find that $K_X^2=0$. If the 9 points are the base points of a pencil of cubics then you find that $h^0(-k K_x)\geq k+1$ for every $k>0$, and Riemann-Roch gives you that $\chi(-k K_X)=1$.

You can write the equation in Weierstrass form and use the point of order 3 to do a 3-descent. Since there are only three bad primes (2,3,7) this seems doable. (This works only if there are no elements of order 3 in the Tate-Shafarevich group).

Ciliberto and Flamini claim that the case $\mathbb{P}^2\to \mathbb{P}^2$ was proven by Kulikov-Kulikov in 2002. (Actually I am surprised that this result has only been proven in this century...). Both proofs are very much characteristic zero proofs.