Can product DFA for intersection of two disjoint languages be minimal?
No.

Counterexample:

Let alphabet $\Sigma = \left\{ 0, 1 \right\}$, languages $L_1 = \left\{ w0 \,\middle|\, w \in \Sigma^\ast \right\}$ (i.e. the last digit is $0$), $L_2 = \left\{ w1 \,\middle|\, w \in \Sigma^\ast \right\}$ (i.e. the last digit is $1$). Note that $L_1 \cap L_2 = \emptyset$.

Then the following DFAs $M_1$ and $M_2$ are minimal for $L_1$ and $L_2$ respectively:





And the cross product of $M_1$ and $M_2$ (for the intersection of languages) is:



But this is not minimal for the empty language. A minimal DFA for the language is:



In fact, cross-product DFA of $M_1$ and $M_2$ recognizes an intersection of languages $L(M_1) \cap L(M_2)$ (if a state of cross-product DFA is accepting when the original two states are both accepting on original DFAs). So in this case a generated cross-product DFA always recognizes the empty language. Since a minimal DFA for the empty language has only one state and cross-product DFA has ((#states of $M_1$) × (#states of $M_2$)) states, almost all cross-product DFAs are non-minimal.

Also, even though if you define a state of cross-product DFA is accepting when either of the original two states is accepting, the above $(L_1, L_2)$ is a counterexample. Since $L_1 \cup L_2 = \Sigma^\ast$, the minimal DFA has only one state.