This is a point I've been confused about proving several times recently, so I'm going to write out the argument in (potentially excessive) detail.

As abx points out, $\phi^*$ gives an injection $H^0(Y, K_Y^{\otimes n}) \hookrightarrow H^0(X, K_X^{\otimes n})$, which implies the stronger statement that $P_n(Y) \leq P_n(X)$ for all $n$, so in particular the Kodaira dimension of $Y$ is at most that of $X$ (the inequality is often strict, e.g. curves of any genus have finite morphisms to $\mathbb{P}^1$). Let's show $n = 1$ for ease of notation, but the argument is identical for all $n$. 

**EDIT:** My definition of Kodaira dimension is that it is the order of growth of the plurigenera, so if $P_n(Y) \leq P_n(X)$ for all $n$, the fact that $\kappa(Y) \leq \kappa(X)$ is immediate. The fact that this is equivalent to the statement about the transcendence degree of the pluricanonical ring, used in Franscesco Polizzi's answer, is not trivial and uses the asymptotic Riemann-Roch theorem.

There is an open set $U \subseteq X$ such that $\phi|_U: U \rightarrow Y$ is an actual morphism (of course, $\phi$ is literally equal to $\phi|_U$ by the definition of a rational morphism, but I want to emphasize that the domain is $U$). Because your varieties are normal and proper, you can ensure that $U$ has codimension at least $2$ (by the valuative criterion and the fact that the local rings of codimension $1$ points are DVR's). By the $S_2$ property of normality, this implies that $K_X \rightarrow \iota_* ((K_X)|_U) = \iota_* K_U$ is an isomorphism, with $\iota: U \hookrightarrow Y$ the inclusion. (I'm using all over the place that $K_S|_V = K_V$ for $V \subseteq S$ with $S$ smooth, which is obvious from the construction at least in the smooth case).

Now, the natural morphism $K_Y \rightarrow (\phi|_U)_* \phi|_U^* K_Y$ is *injective*, because $K_Y$ is a locally free sheaf on the smooth variety $Y$ and $\phi|_U$ is dominant (see [this]
[1] question for a discussion of what could go wrong if you drop the locally free hypothesis). Then, we can compose this with the pushforward of the natural morphism of sheaves $\phi|_U^* K_Y \rightarrow K_U$ to get a morphism of sheaves $K_Y \rightarrow (\phi|_U)_* K_U$. On sections, we get a morphism $H^0(V, K_V) \rightarrow H^0(W, K_W)$ for any dense open $V \subseteq Y$, with $W = (\phi_U)^{-1}(V)$, compatible with restrictions. In particular, this defines the pullback $H^0(Y, K_Y) \rightarrow H^0(U, K_U) = H^0(X, K_X)$. For a dense open $V \subseteq Y$, we have a diagram:

$
\require{AMScd}
\begin{CD}
H^0(Y, K_Y) @>>> H^0(X, K_X)\\
@VVV @VVV \\
H^0(V, K_V) @>>> H(W, K_W)
\end{CD}
$

The vertical arrows are injections because the canonical sheaves are locally free and  everything in sight is an integral variety. Thus, if we can choose some $V$ such that the bottom arrow is injective, the top arrow is automatically injective. 

Now, because you are working over a characteristic $0$ field, a generically finite dominant morphism is *generically étale*. This means that there is a dense open set $V \subseteq Y$ such that $W:= \phi|_U^{-1}(V) \rightarrow V$ is an étale morphism. 

*Aside: this fact is false in positive characteristic (i.e. Frobenius morphisms are everywhere non-étale). On the other hand, I imagine your statement is still true in positive characteristic since inseparable morphisms "don't change anything", but I don't know how to prove this.*

Now, the relative canonical sheaf is zero for étale morphism, so the canonical exact sequence $0 \rightarrow f^* K_T \rightarrow K_S \rightarrow K_{S/T} \rightarrow 0$ for a smooth morphism $f: S \rightarrow T$ shows that the natural morphism $\phi|_U^* K_V \rightarrow K_W$ is an isomorphism. Then, because we already know $K_V \rightarrow (\phi|_U)_* (\phi_U)^* K_V$ is injective, we're done. 





  [1]: http://mathoverflow.net/questions/255981/when-do-surjective-morphisms-induce-injective-maps-on-global-sections-of-coheren "this"