The statement is not true.

**Theorem.** If $\kappa$ is strongly unfoldable and the GCH holds below $\kappa$, then it holds at $\kappa$ also.

This is just because the strongly unfoldable cardinals are $\Sigma_2$-reflecting, and the result also holds for any $\Sigma_2$-reflecting cardinal. The strongly unfoldable cardinals are less than measurable in consistency strength, being equiconsistent with an unfoldable cardinal, and in particular, they are consistent with $V=L$.

Another small instance of the GCH overspill phenomenon, by essentially the same argument, occurs when $\kappa$ is $\Sigma_2$-extendible, which means that $V_\kappa\prec_{\Sigma_2}V_\theta$ for some ordinal $\theta$. If the GCH holds up to $\kappa$, then it will also hold up to $\theta$, and so at $\kappa$ itself. Meanwhile, these $\Sigma_2$-extendible cardinals have no strength at all---they provably exist in ZFC, unless you insist that they are also inaccessible, in which case the strength is strictly weaker than the existence of a Mahlo cardinal.

Meanwhile, your statement is indeed true for many of the large cardinals below measurable. Many of the large cardinal properties below measurable have to do with the existence of size $\kappa$ objects, such as embeddings between $\kappa$-models or homogeneous sets for colorings of size $\kappa$. For any such kind of large cardinal property, if $\kappa$ can be made to have this property indestrucitibly after $\text{Add}(\kappa,1)$, then we can make the GCH fail at $\kappa$ simply by forcing with $\text{Add}(\kappa,\kappa^{++})$. The large cardinal property will be preserved since all the size $\kappa$ objects are added by a size $\kappa$-subforcing, which amounts to $\text{Add}(\kappa,1)$. This is how one can show that it is consistent that the GCH fails first at $\kappa$, when $\kappa$ is weakly compact, ~~indescribable~~, unfoldable, Ramsey, strongly Ramsey and so on. One should simply perform the preparatory forcing to make $\kappa$ indestructible by $\text{Add}(\kappa,1)$, and then appeal to the argument I just gave.

**Update.** I should not have included the indescribable cardinals on the list, because every $\Pi^2_1$-indescribable cardinal is $(\kappa+1)$-strongly unfoldable and so the failure of the GCH at $\kappa$ will reflect down.