It might be helpful to split your question into two parts:
When is it true that if the fibers $X_s \to {\rm Spec} \kappa (s)$ satisfy a certain property for all $s \in S$, $X \to S$ satisfies this property?
When is it true that if the fibers $X_s \to {\rm Spec} \kappa (s)$ satisfy a certain property for all closed $s \in S$, all fibers satisfy this property.
The answer to the first question is almost never positive if you do not make additional assumptions. The two main cases, where I know of a positive answer, are:
Ia. The first one is Francesco's answer: If $X \to S$ is flat (EDIT: and locally of finite presentation) and all fibers are smooth (or etale), then $X \to S$ has the same property. There are variants for other types of singularities.
Ib. If $X \to S$ is proper and all fibers are finite, then $X \to S$ is finite (Grothendieck's version of Zariski's main theorem).
There are variants, as if $f\colon X \to S$ is proper, flat and of finite presentation and $X_s \to {\rm Spec} \kappa(s)$ is an isomorphism for one $s \in S$, then there exists an open neighborhood $U$ of $s$ such that $f^{-1}(U) \to U$ is an isomorphism.
The answer to the second question is much more often positive. There is the following general principle: Assume that $X$ is of finite presentation over $S$. Let $C$ be the set of points $s \in S$ where $X_s \to {\rm Spec} \kappa(s)$ has a certain "decent" property $P$. Usually "decent" should at least imply that you have a property of schemes over a field that is stable under extension of the base field; thus "irreducible" is not a decent property, but "geometrically irreducible" is.
Then $C$ is constructible. $C$ is open if $X$ is in addition proper and flat over $S$.
Of course this is not true for all properties (e.g., for the property of being an affine morphism) but it does hold for many properties. Many statements of this kind are proved in EGAIV, mainly §9 and §12. There is also a rather long list in the appendix of my book with Görtz.
Therefore to answer the second question it suffices to answer the following question.
- Let $C$ be a constructible (resp. open) subset of a scheme $S$ containing the subset $S_0$ of all closed points of $S$. Is $C = S$?
In general the answer is no because there are schemes that do not contain any closed point. If $C$ is constructible, then the answer is positive if $S$ is of finite type over a field or, more generally, if $S$ is Jacobson (e.g. if $S$ is of finite type over a Dedekind ring with infinitely many prime ideals - as the ring of integers). In this case $C \mapsto C \cap S_0$ yields a bijection between constructible subsets of $S$ and constructible subsets of $S_0$.
If $C$ is open, then to get a positive answer it suffices to have a scheme where every point has a closed point in its closure. This is true for quasi-compact schemes (easy to see using that this is certainly true for affine schemes) or locally noetherian schemes (not that easy to see).
Best Answer
No. The blow up of a point on the plane provides a counterexample. You need to add flatness.
Added: It seems I answered something different from what was asked. Perhaps someone can answer the actual question, which isn't so clear to me.
10 seconds later: It looks like Karl Schwede has a counterexample below.