I think the following works. Let $R$ be the commutative ring $k[u_1, u_2]/(u_1 u_2)$. For $j=1$, $2$, let $$A_j = R\langle x_j, y_j \rangle / ( y_j x_j - x_j y_j - u_j,\ u_{3-j} x_j,\ u_{3-j} y_j ).$$ I think, as an $R$-module, $$A_j = R \cdot 1 \oplus \bigoplus_{a+b \geq 1} \left( R/u_{3-j} R \right) \cdot x_j^a y_j^b$$ and that $Z(A_1) = Z(A_2) = R$. Thus, $$A_1 \otimes_R A_2 = $$ $$R \cdot 1 \oplus \bigoplus_{a+b \geq 1} \left( R/u_2 R \right) \cdot x_1^a y_1^b \oplus \bigoplus_{a+b \geq 1} \left( R/u_1 R \right) \cdot x_2^a y_2^b \oplus \bigoplus_{a_1+b_1 \geq 1,\ a_2+b_2 \geq 1} k \cdot x_1^{a_1} y_1^{b_1} x_2^{a_2} y_2^{b_2}.$$ In particular, $Z(A) \otimes_R Z(B)$ is the first summand, $R \cdot 1$, and so $x_1 x_2 \not\in Z(A) \otimes_R Z(B)$.

Then $x_1 x_2$ is central, because it clearly commutes with $x_1$ and $x_2$, and we compute that $[x_1 x_2, y_1] = [x_1, y_1] x_2 = u_1 x_2 = 0$ and $[x_1 x_2, y_2] = x_1 [x_2, y_2] = x_1 u_2 = 0$. More generally, all the monomials $x_1^{a_1} y_1^{b_1} x_2^{a_2} y_2^{b_2}$ for $a_1+b_1$, $a_2 + b_2 \geq 1$ are central in the same way.

I'm not sure what "central $R$-algebra" means, I have been taking it to mean that the natural map $R \to Z(A)$ is an isomorphism. If it just has to be surjective, the OP has already given a solution. I think the following is an answer to my interpretation where we have to have $R = Z(A)$. 

Let $R$ be the commutative ring $k[u_1, u_2]/(u_1 u_2)$. For $j=1$, $2$, let $$A_j = R\langle x_j, y_j \rangle / ( y_j x_j - x_j y_j - u_j,\ u_{3-j} x_j,\ u_{3-j} y_j ).$$ I think, as an $R$-module, $$A_j = R \cdot 1 \oplus \bigoplus_{a+b \geq 1} \left( R/u_{3-j} R \right) \cdot x_j^a y_j^b$$ and that $Z(A_1) = Z(A_2) = R$. Thus, $$A_1 \otimes_R A_2 = $$ $$R \cdot 1 \oplus \bigoplus_{a+b \geq 1} \left( R/u_2 R \right) \cdot x_1^a y_1^b \oplus \bigoplus_{a+b \geq 1} \left( R/u_1 R \right) \cdot x_2^a y_2^b \oplus \bigoplus_{a_1+b_1 \geq 1,\ a_2+b_2 \geq 1} k \cdot x_1^{a_1} y_1^{b_1} x_2^{a_2} y_2^{b_2}.$$ In particular, $Z(A) \otimes_R Z(B)$ is the first summand, $R \cdot 1$, and so $x_1 x_2 \not\in Z(A) \otimes_R Z(B)$.

Then $x_1 x_2$ is central, because it clearly commutes with $x_1$ and $x_2$, and we compute that $[x_1 x_2, y_1] = [x_1, y_1] x_2 = u_1 x_2 = 0$ and $[x_1 x_2, y_2] = x_1 [x_2, y_2] = x_1 u_2 = 0$. More generally, all the monomials $x_1^{a_1} y_1^{b_1} x_2^{a_2} y_2^{b_2}$ for $a_1+b_1$, $a_2 + b_2 \geq 1$ are central in the same way.