It seems to me that a simple example is as follows: Let $R = \mathbb{Z}$, let $A$ be the noncommutative $\mathbb{Z}$-algebra on two generators $x$, $y$, obeying $xy=yx+2$ and let $B = \mathbb{Z}/2 \mathbb{Z}$.

Note that $A$ is basically the Weyl algebra and its center is $\mathbb{Z}$, so $Z(A) \otimes_R Z(B)$ is $\mathbb{F}_2$, but $A \otimes_R B$ is $\mathbb{F}_2[x,y]$.

It seems to me that a simple example is as follows: Let $R = \mathbb{Z}$, let $A$ be the noncommutative $\mathbb{Z}$-algebra on two generators $x$, $y$, obeying $xy=yx+2$ and let $B = \mathbb{Z}/2 \mathbb{Z}$.

Note that $A$ is basically the Weyl algebra and its center is $\mathbb{Z}$. Here is a more detailed argument: Let $W = \mathbb{Q}\langle x,y \rangle / (xy-yx-2)$; $W$ is isomorphic to the Weyl algebra. There is an obvious map $A \to W$. We claim that it is injective. Proof: By a PBW like argument, $x^i y^j$ spans $A$ over $\mathbb{Z}$, and these monomials are linearly independent in $W$. $\square$ So $x^i y^j$ is a $\mathbb{Z}$-basis of $A$, and we can compute in it in the usual way to see that $\mathbb{Z} = Z(A)$.

So $Z(A) \otimes_R Z(B)$ is $\mathbb{F}_2$, but $A \otimes_R B$ is $\mathbb{F}_2[x,y]$.