The maximum possible number of moves is $\max(ab, ac, bc)$.
First, we prove this is best possible. We define a feisty triplet to be an unordered triple of ducks, one of each of rock, paper, scissors, such that the paper duck is between the rock and scissors duck and facing the rock duck, as shown. (There may be other ducks not pictured, but the orders are irrelevant.)

Claim — The number of feisty triplets decreases by $c$ if a paper duck swaps places with a rock duck, and so on.
Proof. Clear. ☐

Obviously the number of feisty triples is at most $abc$ to start. Thus at most $\max(ab, bc, ca)$ moves may occur, since the number of feisty triplets should always be nonnegative, at which point no moves are possible at all.
To see that this many moves is possible, assume WLOG $a = \min(a, b, c)$ and suppose we have $a$ rocks, $b$ papers, and $c$ scissors in that clockwise order.

Then, allow the scissors to filter through the papers while the rocks stay put. Each of the $b$ papers swaps with $c$ scissors, for a total of $bc = \max(ab, ac, bc)$ swaps.