Yes. At first, if $h=\min(h_A,h_B)$ is convex (note that it is also 1-homogeneous), it is a support function of the body $C:=\{x:\forall\xi\in \mathbb{R}^n,\langle \xi,x\rangle\leqslant h(\xi)\}$. Next, $C=A\cap B$, since the inequality $\langle \xi,x\rangle\leqslant h(\xi)$ is equivalent to a system of two inequalities $\langle \xi,x\rangle\leqslant h_A(\xi)$, $\langle \xi,x\rangle\leqslant h_B(\xi)$. Now assume that $A\cup B$ is not convex. It means that there exist $a\in A$, $b\in B$ such that the segment between $a$ and $b$ is not covered by $A\cup B$. Look at a line between $a$ and $b$, it intersects $A$ by a segment $[a_1,a_2]$ and $B$ by a segment $[b_1,b_2]$, we may suppose that $a_1<a_2<b_1<b_2$ on this line. Consider two convex compact sets: $[a_2,b_1]$ and $C=A\cap B$. They are disjoint, thus separated by a hyperplane. In other words, there exists $\xi\in \mathbb{R}^n$ such that $\sup_{x\in C} \langle \xi,x\rangle<\inf_{x\in [a_2,b_1]} \langle \xi,x\rangle$. It follows that $h(\xi)\geqslant \min(\langle \xi,a_2\rangle,\langle \xi,b_1\rangle)>h_C(\xi)$, a contradiction.