Dear kaddar, here is a partial answer.

According to a theorem of Douady, a flat map $f:X\to S$ between complex analytic spaces is always open . So if you assume that the fibers of $f$ are reduced *and* that your reduced space $S$ is actually smooth (i.e. is a manifold), then $X$ is indeed reduced : this follows from the Proposition on page 158 of Gerd Fischer's *Complex Analytic Geometry*
(Springer, LNM 538, 1976).

**Edit** $\;$ On the evoked relation between flat and open, let me add the following. It is *not* true that an open morphism $f:X\to S$ of complex spaces is flat: the simplest counter-example is the immersion of a simple point into a double point i.e. the morphism of schemes $Spec \;\mathbb C \to Spec \; \mathbb C[\epsilon ] \quad(\epsilon^2=0)$ seen analytically.
However if $X$ and $S$ are complex manifolds then it *is* true that $f$ open implies $f$ flat (Fischer, same page 158)
and so for morphisms between manifolds you have the easy to remember equivalence flat=open, which helps understand the notoriously unintuitive notion of flatness.