Does the functor Sch to Top have a right adjoint?

Another way to see that the functor $\mathrm{Sch} \to \mathrm{Top}$ is not a left adjoint is to see that it does not preserve colimits. In this MO answer, Laurent Moret-Bailley gives an example of a pair of arrows $Z \rightrightarrows X$ in $\mathrm{Sch}$, such that the canonical map from $X$ to the coequalizer $Y$ is not surjective (as a function between the sets of points of the underlying spaces). Since in $\mathrm{Top}$ those canonical maps to the coequalizer are always surjective, this coequalizer cannot be preserved by the forgetful functor.

No such right adjoint exists, even restricted to sober spaces. For simplicity let us take $S=\operatorname{Spec} k$ for some field $k$, and consider the space $X$ having two points, one of which is closed. If $G(X)$ existed, then maps $M\to G(X)$ would be in bijection with closed subsets of $M$. It is not hard to show no such $G(X)$ exists. For instance, taking $M$ to be Specs of fields extending $k$, you can see $G(X)$ must only have two points, and in particular it must be affine. You then get a $k$-algebra $A$ with a radical ideal $I$ such that for any $k$-algebra $B$ with a radical ideal $J$, there is a unique map $f:A\to B$ such that $J$ is the radical ideal generated by $f(I)$. Clearly no such $(A,I)$ can exist, since for any cardinal $\kappa$ we can find a $(B,J)$ such that $J$ cannot be generated by fewer than $\kappa$ elements.