The following is a bit more elementary (and purely algebraic) in nature than the previous examples. Take $A=B=\mathbb{C}^2$. Identify $A\otimes B\cong\mathbb{C}^4$. Observe that for trivial linear algebra reasons, any tensor product representation of $A\otimes B$ generates a C-algebra in its range that has dimension 0, 1, 2, or 4. Take $H_1=\mathbb{C}^2=H_2$ and consider any representation that generates a $3$-dimensional C-algebra, such as $$ \pi: A\otimes B\cong\mathbb{C}^4\to B(H_1\otimes H_2)\cong M_4\otimes M_4, \pi(\lambda_1,\lambda_2,\lambda_3,\lambda_4) = \operatorname{diag}(\lambda_1,\lambda_1,\lambda_3,\lambda_4)\otimes 1. $$ Alternatively, $H_1=\mathbb{C}^2, H_2=\mathbb{C}$ also does the trick via $$ \pi: \mathbb{C}^2\otimes\mathbb{C}^2\to M_2, \pi(e_i\otimes e_j) = \operatorname{diag}(\delta_{1,i},\delta_{2,j}). $$ Although the image here has dimension 2, a small calculation shows that this cannot be of product form, either.

The following is a bit more elementary (and purely algebraic) in nature than the previous examples. Take $A=B=\mathbb{C}^2$. Identify $A\otimes B\cong\mathbb{C}^4$. Observe that for trivial linear algebra reasons, any tensor product representation of $A\otimes B$ generates a C-algebra in its range that has dimension 0, 1, 2, or 4. Take $H_1=\mathbb{C}^2=H_2$ and consider any representation that generates a $3$-dimensional C-algebra, such as $$ \pi: A\otimes B\cong\mathbb{C}^4\to B(H_1\otimes H_2)\cong M_2\otimes M_2 \cong M_4, \pi(\lambda_1,\lambda_2,\lambda_3,\lambda_4) = \operatorname{diag}(\lambda_1,\lambda_1,\lambda_3,\lambda_4)\otimes 1. $$ Alternatively, $H_1=\mathbb{C}^2, H_2=\mathbb{C}$ also does the trick via $$ \pi: \mathbb{C}^2\otimes\mathbb{C}^2\to M_2, \pi(e_i\otimes e_j) = \operatorname{diag}(\delta_{1,i},\delta_{2,j}). $$ Although the image here has dimension 2, a small calculation shows that this cannot be of product form, either.