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 answer is no.
The reason is that 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? Furthermore, 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 $g\ast s$ made $\kappa$ into a singular cardinal without collapsing any cardinals. This combined forcing has size $\kappa^+$ under the GCH.
Although $\kappa$ is not measurable in $V$, it was measurable in an inner model of $V$. This leads to another question that is closely related to 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 it seems likely that one might get $0^\sharp$ and 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.