The intersection of all subsets of an ultrafilter, if it is nonempty, must be a singleton; therefore a free ultrafilter is the same as a non-principal ultrafilter. It follows that a uniform ultrafilter on an infinite set, because it cannot contain a singleton, must be free. (The only finite set that has a uniform ultrafilter is the one-element set; we won't count that.)
Equivalently, a free ultrafilter is one which contains the filter of cofinite sets. This means a free ultrafilter cannot contain a finite set, so if the given set is countable, then a free ultrafilter on that set is uniform.
However, a free ultrafilter on an uncountable set $X$ need not be uniform (if we accept that every proper filter can be extended to an ultrafilter, which follows from the axiom of choice). For if $C \subseteq X$ is cofinite and $A \subseteq X$ is some given countable subset, then we cannot have $A \cap C = \emptyset$ (for that would imply $A$ is contained in the complement of $C$, hence contained in a finite set). It follows that sets of the form $A \cap C$, where $C$ ranges over cofinite sets, generate a proper filter, which is contained in some ultrafilter. This ultrafilter would contain $A$ and $X$, hence two sets of different cardinalities.