Yes

Here is a winning strategy :

First play Paper. Just because.

Then at each move there are 3 possibilities:

If you have just made a draw (wlog, you both played Rock) then you both have the same possibilities for the next move, one beating the other (here Scissors > Paper). Play this one (here Scissors) with an expected score of 3/4.

If you have just lost (wlog, you played Rock against Paper), then one of your two possibilities can make either a draw or a loss (here Scissors, vs Scissors or Rock) and the other can either win or lose (here Paper, vs Scissors or Rock). Play the latter one with an expected score of 1/2.

If you have just won (wlog, you played Paper against Rock), then one of your two possibilities can make either a draw or a win (here Scissors, vs Scissors or Paper) and the other can either win or lose (here Rock, vs Scissors or Paper). Play the former one with an expected score of 3/4.

Your expected win is

a linear combinaison of 3/4, 1/2 and 3/4 with non-zero probabilities, so it is higher than 1/2.

Your probablities for round n (n>1) are:

loss : P_L(n) = 1/3 * 2^(1-n)
draw : P_D(n) = 1/3 * (2^(n-1)-1) * 2^(1-n)
win : P_W(n) = 1/3 * (2^(n-1)-1) * 2^(1-n)

When the number of games tends to infinity, your expected score

tends to 3/4.