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 ultrafilteron that set.
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.
