Your question is very interesting in the case $\kappa^+<2^\kappa$.

For this case, recall that a cardinal $\kappa$ is weakly measurable if every family of $\kappa^+$ many subsets of $\kappa$ admits a $\kappa$-complete nonprincipal filter measuring them. This notion was introduced and studied by my student Jason Schanker in his dissertation. There are a variety of equivalent characterizations.

Jason proved that if $2^\kappa$ is larger than $\kappa^+$, then this concept is not necessarily equivalent to measurability, for he constructed models of ZFC with a weakly measurable cardinal, but no measurable cardinal.

In such a model, therefore, where $2^\kappa=\kappa^{++}$, we have measures for all the families of sets of size smaller than $2^\kappa$, but $\kappa$ is not measurable. So this provides a negative answer to your question.

Meanwhile, weakly measurable cardinals are necessarily weakly compact, inaccessible and much more, so there is a positive answer to the version of your question at the end where you ask for less than measurability.

Indeed, the existence of $\kappa$-complete nonprincipal filters measuring families of $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact. In this sense, weak measurability is a generalization of weak compactness from $\kappa$ to $\kappa^+$.

Thus, if $2^\kappa=\kappa^+$, the answer to your question is no, since the existence of such filters for families of at most $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact, which is a strictly weaker large cardinal concept than measurability.