A pair of optimal pants decompositions $A, B$ 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 $4\cosh^{-1}(\sqrt{2}) = 3.525494\ldots$. QED

Thus both $A$ and $B$ are optimal, yet every curve of $A$ crosses every curve of $B$.