If rational and mod $p$ cohomology vanish above a certain degree, does integral cohomology also vanish?

If you're working with spaces having f.g. cohomology, the result in the post (not the title) holds. All $\text{Ext}(H_n(X, \mathbb{Z}), \mathbb{Z}_p)$ vanish by the universal coefficient theorem, so $H_n(X, \mathbb{Z})$ contains no torsion. It follows from the universal coefficient theorem for homology that $H_{n+1}(X, \mathbb{Z})$ must vanish, and so $H^{n+1}(X, \mathbb{Z})$ must too.

Almost but not quite. If $H^*(X;F)=0$ for $*>n$ and all prime fields $F$, then by the universal coefficient theorem, $\operatorname{Hom}(H_*(X;\mathbb{Z}),F)=\operatorname{Ext}(H_*(X;\mathbb{Z}),F)=0$ for $*>n$. But if $\operatorname{Hom}(A,\mathbb{Q})=0$ then $A$ must be torsion, and any injection $\mathbb{Z}/p\to A$ induces a surjection $\operatorname{Ext}(A,\mathbb{Z}/p)\to\operatorname{Ext}(\mathbb{Z}/p,\mathbb{Z}/p)\neq0$.

It follows that we must have $H_*(X;\mathbb{Z})=0$ for all $*>n$, and so $H^*(X;\mathbb{Z})=0$ for $*>n+1$ and $H^{n+1}(X;\mathbb{Z})=\operatorname{Ext}(H_n(X;\mathbb{Z}),\mathbb{Z})$. However, this latter group could be nontrivial: for instance, $H_n(X;\mathbb{Z})$ could be $\mathbb{Q}$, in which case $H^{n+1}(X;\mathbb{Z})=\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$ is uncountable.