If you are over a field of char. $0$, then yes, it is true that the fibre over a generic point of the image is reduced. To see this, first replace $W$ with the closure of the image of $\phi$, so as to assume that $\phi$ is dominant. Then the morphism $\phi$ corresponds to an injection of finite-type domains over $k$, say $A \hookrightarrow B$. (The injectivity follows from the dominance of $\phi$.) Now let's look at the generic fibre of $\phi$: this is Spec of the tensor product $K(A) \otimes_A B$ (here $K(A)$ denotes the fraction field of $A$), which is a localization of $B$, and so a domain (and hence reduced). Now if we are in char. zero, the reduced $K(A)$-algebra $K(A)\otimes_A B$ is necessarily geometrically reduced (i.e. stays reduced after extending to an algebraic closure of $K(A)$), and the property of fibres being geometrically reduced is open on the base, i.e. on Spec $A$; thus the fibres over an open subset of the image of $\phi$ will be reduced. (See e.g. Thm. 2.2 of these notes. You can find this in many places (though mabye not Hartshorne); these notes are just what turned up near the top of a quick google search.)
In char. $p$, this need not be true. E.g. consider the Frobenius map $\mathbb A^1 \to \mathbb A^1$, given by $x \mapsto x^p$. Then the fibre over every (closed) point is non-reduced. (The fibre over the generic point is reduced, but not geometrically reduced.) Nevertheless, if the generic fibre (in the scheme-theoretic sense) is geometrically reduced, then the fibre over an open set of closed points will be reduced too.