If your filter is generated by $\kappa$ many sets, then indeed the
conclusion you seek can be made, by a direct argument that does
not go through strong compactness.
Theorem. The following are equivalent, for any uncountable
regular cardinal $\kappa$.
- $\kappa$ is a measurable cardinal.
- Every $\kappa$ complete filter $F$, generated by at most
$\kappa$-many sets, extends to a $\kappa$-complete ultrafilter.
Proof: It is easy to see that $2$ implies $1$, since the filter of
co-bounded sets in $\kappa$ is $\kappa$-complete and generated by
the tails, so there is a $\kappa$-complete non-principal
ultrafilter on $\kappa$.
For the main direction, assume $\kappa$ is measurable and $F$ is a
$\kappa$-complete filter on a set $D$ with $F$ generated by at
most $\kappa$ many sets $X_\alpha$, for $\alpha\lt\kappa$. Let
$j:V\to M$ be an elementary embedding with critical point
$\kappa$. By applying $j$ to $\vec X=\langle
X_\alpha\lt\kappa\rangle$ and restricting to $\kappa$, we see that
$\langle j(X_\alpha)\mid\alpha\lt\kappa\rangle$ is in $M$. And
since this is fewer than $j(\kappa)$ many elements of $j(F)$,
which is $j(\kappa)$-complete in $M$, it follows that
$\bigcap_{\alpha\lt\kappa}j(X_\alpha)\in j(F)$, and in particular,
there is some $a\in \bigcap_\alpha j(X_\alpha)$. Define
$U=\{X\subset D\mid a\in j(X)\}$. It is easy to verify that $U$
is a $\kappa$-complete ultrafilter on $D$ and $F\subset U$, as
desired. QED
For $\theta$-generated filters, one generally needs
$\theta$-strong compactness, as mentioned in the comments, and this is in fact equivalent to $\theta$-strong compactness. The essence of the argument above, then, is that a cardinal $\kappa$ is measurable if and only if it is $\kappa$-strongly compact.
That said, if you want this filter extension property, then I encourage you to go ahead and make the strong compactness assumption. There are many beautiful theorems using strongly compact cardinals.
