Optimal pants decompositions need not have disjoint shortest curves. Here is an example.
Let $S$ be the genus two hyperbolic surface built from four equilateral right-angled hexagons by "doubling". That is, let $H$ be such a hexagon. Let $a = 2 \cosh^{-1}(\sqrt{3/2}) = 1.31695\ldots$ denote the side-length of $H$. (See Beardon.) Let $\alpha$ be the union of three non-adjacent sides of $H$ and let $\beta$ be the other three non-adjacent sides. Doubling $H$ across $\alpha$ gives a pants $P$. So $\alpha$ gives the seams of $P$ and $\beta$ doubles to give $\partial P$, each component having length $2a$. Now double across $\beta$ to get $S$.
Let $B = \partial P$ be the double of $\beta$ in $S$. So $B$ is a pants decomposition of $S$, each curve having length $2a$. Likewise $A$, the double of $\alpha$, is a pants decomposition.
Claim: All curves of $A$ and $B$ are systoles of $S$.
Proof: Any geodesic in a genus two surface is preserved by the hyperelliptic and so double covers an arc or a loop in the orbifold $O = S^2(2,2,2,2,2,2)$. In our situation $O$ is the double of $H$ across $\partial H$. The shortest arc is an edge of $H$. The shortest loop divides $O$ in half and has length $4\cosh^{-1}(\sqrt{2)}) = 3.5254943480$. QED
Thus both $A$ and $B$ are optimal, yet every curve of $A$ crosses every curve of $B$.