The answer is yes, and the morphism doesn't have to be smooth.
A fancy way to explain this
is that if $f:W\to S$ is a morphism of complex algebraic varieties, $f^*:H^n(S)\to H^n(W)$
is compatible with mixed Hodge structures (by Deligne), and the functor
$H\mapsto H^{pq}:= Gr^p_FGr_{p+q}^WH$ form MHS to vector spaces is exact
In the case that I suspect you actually care about when $W$ and $S$ are both smooth and projective, there is an easier way to see this. In this case, there is canonical Hodge structure determined
by the Hodge filtration: $\alpha\in F^pH^n(S)$ if $\alpha$ is represented by a sum of
forms of type $(p',q')$ with $p'\ge p$. One recovers the usual bigrading by $H^{pq}(S)=F^p\cap \bar F^q H^n(S)$ and $f^*$ clearly preserves the bigrading.
So your desired statement follows by decomposing $f^*\alpha$ into $(p,q)$-type.