Since the axioms describing what a valuation ring can be put as what's called *geometric sequents* [*], by the fundamental theorem on classifying toposes, there is a topos $T_{val}$ with precisely the universal property you're asking for. (See for instance Section 2.1.2 and more specifically Theorem 2.1.8 in Olivia Caramello's book *Theories, Sites, Toposes*.)

But then there is the additional question whether we can recognize this topos as one of the toposes commonly used in algebraic geometry, just as we can recognize the classifying topos of local rings as the big Zariski topos of $\operatorname{Spec}(\mathbb{Z})$.

As far as I know, this question is open. The answer linked to by Robert in the comments gives some indication that $T_{val}$ might coincide with the big rh topos of $\operatorname{Spec}(\mathbb{Z})$, but just that $T_{val}$ and the big rh topos have the same topos-theoretic points, as set out in the linked answer, doesn't yet mean that the toposes are equivalent.

You could also ask the analogous question for algebraically closed valuation rings. In this case again a classifying topos exists, by the same abstract nonsense, and
conditional on some representability conjecture it coincides with the big ph topos of $\operatorname{Spec}(\mathbb{Z})$ (see Proposition 21.18 in these notes of mine).

[*] A geometric sequent is a logical formula of the form $\forall \cdots \forall. (\cdots) \Rightarrow (\cdots)$ where in the two bracketed subformulas the connectives $\Rightarrow$ $\neg$ $\forall$ may not occur. For instance the field axiom "$\forall x : R. x = 0 \vee (\exists y : R. xy = 1)$" can be put as a geometric sequent (just imagine an invisible "$\top \Rightarrow$"), but the (classically but not intuitionistically) equivalent axiom "$\forall x : R. \neg(\exists y : R. xy = 1) \Rightarrow x = 0$" cannot.