Suppose for contradiction that Bob has a winning strategy.
Claim: If Alice makes the first move on square X, then on the next turn (Bob's), there is exactly one square Y among the neighbors of X on which Bob must make his (first) move to win.
Assuming the claim, for each square X, let f(X) denote the Y as in the claim. Then since there are more W squares than B, there must be two W squares $X_1,X_2$ and a B square Y such that $f(X_1)=f(X_2)=Y$.
Now this gives a strategy for Alice to win by starting at $Y$: if Bob places the knight on a square other than $X_1$, Alice steals Bob's strategy by pretending that Bob already made a move at $X_1$; if Bob places the knight on $X_1$, Alice pretends that the first move was Bob's at $X_2$ and again wins by strategy-stealing.
Proof of the claim:
Suppose for contradiction that Y, Z are two squares both adjacent to X and such that when Alice starts at X, then Bob has a winning strategy by placing his knight at Y or at Z.
Let Alice begin a game by placing the knight at Y.
Now if Bob doesn't place the knight at X, then Alice can pretend to be the second-player where the game began with Bob starting at X; this is a contradiction so Bob must place the knight at X.
But in this case Alice can place the knight on Z and by strategy-stealing again, go on to win. This proves the claim.