Let $\alpha_0<\alpha_1$ be the least ordinals (in the reverse lex order, say) such that $V_{\alpha_0}$ and $V_{\alpha_1}$ have the same second order theory $T$. Assume towards a contradiction that $V_{\alpha_0}$ and $V_{\alpha_1}$ are elementarily embeddable into a common structure $M$, which we may assume is transitive. Since $V_{\alpha_1}$ satisfies that there is an ordinal $\beta$ such that $V_\beta$ satisfies $T$, so does $M$, and hence so does $V_{\alpha_0}$. But this means that some $\beta < \alpha_0$ has the same theory as $V_{\alpha_0}$, contrary to the minimality of the pair $\alpha_0 < \alpha_1$.

The answer to the question is yes.

Let $\alpha_0<\alpha_1$ be the least ordinals (in the reverse lex order, say) such that $V_{\alpha_0}$ and $V_{\alpha_1}$ have the same second order theory $T$. Assume towards a contradiction that $V_{\alpha_0}$ and $V_{\alpha_1}$ are elementarily embeddable into a common structure $M$, which we may assume is transitive. Note that the embeddings fix $T$. Since $V_{\alpha_1}$ satisfies that there is an ordinal $\beta$ such that $V_\beta$ satisfies $T$, so does $M$, and hence so does $V_{\alpha_0}$. But this means that some $\beta < \alpha_0$ has the same theory as $V_{\alpha_0}$, contrary to the minimality of the pair $\alpha_0 < \alpha_1$. (Edit: I now see this is very close to what Trevor was doing.)