Your question is asking whether the measurability of a measurable cardinal is downwards absolute to ground models: if $\kappa$ is measurable in a forcing extension $V[G]$, must it be measurable in $V$? This is a question that makes sense for any of the large cardinal notions.
The answer is that, although the smaller large cardinal notions such as inaccessibility and Mahlo-ness are downward absolute, this phenomenon does not generally hold for the much larger large cardinals.
Here is one way to see it. Suppose $\kappa$ is a measurable cardinal in $V$. We may force $2^\kappa=\kappa^+$ while preserving the measurability of $\kappa$, since this adds no new subsets to $\kappa$; so suppose $2^\kappa=\kappa^+$ already in $V$. Let $\mathbb{P}$ be the Easton support forcing iteration of length $\kappa$, which forces with $\mathbb{Q}_\gamma=\text{Add}(\gamma,1)$ to add a Cohen subset at every inaccessible cardinal $\gamma$. Let $\mathbb{Q}_\kappa=\text{Add}(\kappa,1)$ be the stage $\kappa$ forcing, to do so at the top. Suppose $G\ast g\subset\mathbb{P}\ast\mathbb{Q}_\kappa$ is $V$-generic. First, I claim that $\kappa$ is measurable in $V[G][g]$. This follows from the usual lifting arguments, which appear in many of my papers. Start with $j:V\to M$, an ultrapower embedding by a measure on $\kappa$. The forcing $j(\mathbb{P})$ is isomorphic to $\mathbb{P}\ast\mathbb{Q}\ast\mathbb{P}_{tail}$, where the tail forcing is $\leq\kappa$-closed in $M[G][g]$. Since $|j(\kappa^+)|^V=\kappa^+$, we may enumerate the dense subsets of $\kappa$ in $M[G][g]$ in a $\kappa^+$ sequence in $V[G][g]$. And since $M[G][g]^\kappa\subset M[G][g]$ in $V[G][g]$, we may thereby diagonalize to produce an $M[G][g]$-generic filter $G_{tail}\subset\mathbb{P}_{tail}$, and thus lift $j$ to $j:V[G]\to M[j(G)]$, where $j(G)=G\ast g\ast G_{tail}$. Similarly, the object $g$ is essentially a condition in $j(\mathbb{Q}_\kappa)$, and so we may diagonalize again to produce an $M[j(G)]$-generic filter $h\subset j(\mathbb{Q}_\kappa)$ containing it, and thus lift $j$ fully to $j:V[G][g]\to M[j(G)][j(g)]$, with $j(g)=h$. Thus, $\kappa$ is measurable in $V[G][g]$.
Meanwhile, and this is the main point, I claim that $\kappa$ is not measurable in $V[G]$. Suppose it were, with embedding $j:V[G]\to \bar M$ in $V[G]$. By elementarity, since $V[G]$ thinks it is a forcing extension by $\mathbb{P}$, it follows that $\bar M=M[j(G)]$ for some inner model $M$. (My theorems on approximation and covering show that in fact $M\subset V$ and $j\upharpoonright V:V\to M$ is definable in $V$, but we don't need that here.) We may factor the forcing as $M[j(G)][G\ast \bar g\ast\bar G_{\rm tail}]$, where $g'\subset\text{Add}(\kappa,1)^{M[G]}$ is $M[G]$-generic. But note that $P(\kappa)^V\subset M$ and so also $P(\kappa)^{V[G]}\subset M[G]$. Thus, $g'$ will have to be really $V[G]$-generic for this forcing, which is a contradiction if $g'\in V[G]$. So there can be no such embedding in $V[G]$, and so $\kappa$ is not measurable there.
Thus, we have a model $\bar V$, namely $\bar V=V[G]$, where $\kappa$ is not measurable, but it becomes measurable after forcing with $\text{Add}(\kappa,1)$.
The same essential argument works with all the stronger large cardinals. If we had started with $\kappa$ supercompact, for example, then we could make a model where it is not measurable, but becomes supercompact after forcing with $\text{Add}(\kappa,1)$.
Kunen observed that one can make a more extreme example as follows (Saturated ideals, {\em Journal of Symbolic Logic}, 43(1):65--76, March 1978). He relies on the fact that if one first adds a homogeneous Souslin $\kappa$-tree, and then forces with that tree, the combined forcing is equivalent to $\text{Add}(\kappa,1)$. But the first step kills the weak compactness of $\kappa$. So combining this observation with the previous, one arrives at the conclusion:
If $\kappa$ is a measurable cardinal, then there is a forcing extension $\bar V$ in which $\kappa$ is no longer weakly compact, but forcing with a $\kappa$-Souslin tree in $\bar V$ makes $\kappa$ suddenly measurable in the forcing extension $\bar V[g]$. Furthermore, if $\kappa$ was tall, strong or supercompact in th original ground model, then it will also retain those stronger properties in $\bar V[g]$.
So a non-weakly compact cardinal can become supercompact by forcing.
