Let $p_i$ be a list of the rational numbers. Let $U_{i,n}$ be an open interval centered on $p_i$ of length $2^{-i}/n$. Then $V_n=\cup_i U_{i,n}$ is an open cover of the rationals, of measure at most $\sum_i 2^{-i}/n=2/n$. Then $\cap_n V_n$ is a co-meager set of measure zero.

Let $p_i$ be a list of the rational numbers. Let $U_{i,n}$ be an open interval centered on $p_i$ of length $2^{-i}/n$. Then $V_n=\cup_i U_{i,n}$ is an open cover of the rationals, of measure at most $\sum_i 2^{-i}/n=2/n$. Then $\cap_n V_n$ is a co-meager set of measure zero.

So yes, there is a measure zero set that is not meager, and so no, not every measure zero set is meager.

Computability theory gives a neat way to look at this. There is a certain type of real number that is called 1-generic and there is another type that is called 1-random or "Martin-Löf random". These two sets are disjoint. The set of 1-generic reals is co-meager and has measure zero, whereas the set of 1-random reals is meager and has full measure.

Thus measure and category are quite orthogonal. Set theorists would say they correspond to two different notions of forcing.

A good general reference for this kind of question is Oxtoby's classic book Measure and category.