Definitely no. To see why, consider just the case where $W=\mathbb Q_\ell$ is the trivial representation (this is not really a loss of generality, because extensions of $W$ by $V$ are "the same thing" as extensions of $\mathbb Q_\ell$ by $W^\ast \otimes V$). Then extensions (as $G$-representations) of $\mathbb Q_\ell$ by $V$ are parametrized by $H^1(G,V)$. You're asking if such an extension which is
split at almost all places $w$ (say all places $w$ not in a finite set $S$) is split. In other words, you're asking if the
kernel $H^1_S(G,V)$ of $H^1(G,V) \mapsto \prod_{w \not \in S} H^1(G_w,V)$ is trivial.

The spaces $H^1_S(G,V)$ are examples of Selmer groups and while they are known to be finite dimensional, they are not always 0. In fact, computing their dimensions in general is still an open problem, essentially the characteristic zero part of the Bloch-Kato conjecture.

To give an example where $H^1_S(G,V)$ is not zero, let $K=\mathbb Q$, $V=\mathbb Q_\ell(n)$ with $n \neq 1$, $n$ odd (that is $V$ is the one-dimensional rep. where $G$ acts as the cyclotomic character to the power $n$) and $S$ a finite set of places containing $\ell$. Then it is known that $H^1(G,V)$ has dimension 1 in this case (to prove that it has dimension at least 1, which is enough for the argument, use Global Poitou-Tate Euler-Poincaré. For at most 1, you need deep result of Soulé). On the other hand, $H^1(G_w,V)$ has dimension $0$, as follows easily from local Euler-characteristic formula. Hence $H^1_S(G,V)$ has dimension 1. In other words, there is one (and only one) non-split extension of $1$ by $\mathbb Q_\ell(n)$, but it is split at all places $w \neq \ell$. That's a counter-example to your question. (It is interesting to note that the question whether this extension is split as $\ell$ or not is still open for positive $n$).

Let me add (paragraph removed after Kevin's comment)