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.