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Yes, since $$\phi_*K_X=K_Y.$$ Here is the reason. We take a common resolution of $X$ and $Y$, say $$ p:W\rightarrow X, \\q: W \rightarrow Y. $$ Then we can right $$ K_W-p^*K_X=E,\\ K_W-q^*K_Y=F, $$ where $E$ is exceptional over $X$ (and over $Y$ because "contraction") and $F$ is exceptional over $Y$. If apply $q_*$ to $$ p^*K_X+E=q^*K_Y=F, $$ we get the equality.