Is there a winning strategy for this number game?

Let $W=\{n\in Z_{>2}|\;\; 2|n \wedge \exists p\in\mathbb P_{>2}: p|n \}\cup\{4\}$. Say it's your turn on the position $n\in W$. If you select a move $n-p$ where $p>2$ then the opponent is on an odd number $n-p\notin W$ and must select to subtract an odd prime $q$, where $p|(n-p)$ and $q|(n-p-q)$. But $(n-p-q)\in W$ since it's even with a prime factor $>2$.

By induction you may stay in a path in $W$ and as your number get smaller you will get a position $2r$ where $r\in\mathbb P_{>2}$ and win with the move $2r-r$, if and only if the opponent's odd number doesn't reach a prime befor you do. But the number $n-p-q$ is a prime only if $n-p-q=q$ (since $q|(n-p-q)$), that is if $n-p=2q$ which is impossible since $n-p$ is odd.

There is no winning strategy on an odd number. From an odd number one have to go an even number that's not a power of two. The other player then have a position in $W$ which is possible to keep while subtracting an odd prime from the even number, giving the opponent a position at an other odd non prime number.

Example for start 15 and 20
Losing position: 
15,10,5
   12,9,6,3
Winning position:
20,15 se above.