Let me answer the question that I believe you are trying to ask. Namely, if we can make a regular cardinal $\kappa$ into a singular cardinal $\kappa$ by forcing of size at most $\kappa^+$, without collapsing any cardinals, must $\kappa$ be measurable?
The question is very natural, since Prikry forcing is the main way to do something like that, but it requires a measurable cardinal. Nevertheless, the answer is no.
The reason is that we can have a non-measurable cardinal that becomes measurable, and so the combined forcing of first making it measurable and then using Prikry forcing can exhibit your features. Specifically, it is consistent with ZFC (relative to the existence of a measurable cardinal) that there is a non-measurable cardinal $\kappa$ that becomes measurable in a forcing extension, by forcing to add a Cohen subset to $\kappa$. This is explained in my answer to Trevor Wilson's question Can measures be added by forcing? One can arrange in that argument that the GCH holds and that there are no other measurable cardinals.
So suppose that $V$ satisfies ZFC+GCH and there are no measurable cardinals in $V$, but $\kappa$ becomes measurable in $V[g]$, where $g$ was $V$-generic for the forcing to add a Cohen set $g\subset\kappa$. This does not collapse cardinals. Since $\kappa$ is measurable in $V[g]$, we may now perform Prikry forcing over $V[g]$ to add a Prikry sequence $s$, which changes the cofinality of $\kappa$ to $\omega$, while preserving all cardinals.
So in $V$, there were no measurable cardinals and $\kappa$ was regular, but the combined forcing to add $g\ast s$, forcing which has size $\kappa^+$ under the GCH, made $\kappa$ into a singular cardinal without collapsing any cardinals. Thus, this is a counterexample to the requested implication.
Meanwhile, although $\kappa$ is not measurable in $V$, it was measurable in an inner model of $V$, and this leads naturally to a closely related version of your question:
Question. If we can force a regular cardinal $\kappa$ to be singular with forcing of size at most $\kappa^+$ and without collapsing any cardinals, must there be an inner model with a measurable cardinal?
I don't know without further thought (although I recall having had conversations about this question). It seems likely that one might get $0^\sharp$ and perhaps much more out the hypothesis by combining the forcing with a collapse of $\kappa^+$, which would violate Jensen's theorem. We may have to wait for the inner model theory experts.