Yes, I think taking invariants is exact in your context.

**Proof:** Let be $G$ a pro-p group and $0 \to A \to B \to C\to 0$ a short exact sequence (s.e.s) of finite dim. $\mathbb{Q}_l$-vector spaces on which $G$ acts continuously and $\mathbb{Q}_l$-linearly. There is a long exact sequence (l.e.s.) [N, 2.3.2]

$$0 \to A^G \to B^G \to C^G \to H^1_c(G,A) \to H^1_c(G,B) \to \cdots$$

of $\mathbb{Q}_l$-vector spaces ($H^1_c$ benotes continuous cohomology). Hence it's enough to show $H^1_c(G,A)=0$.

Write $A= \mathbb{Q}_l^n$. By the OP's comment above, there is a $G$-submodule $T = \mathbb{Z}_l^n$ of $A$ and $G$ acts continously and $\mathbb{Z}_l$-linearly. With $W := A/T = (\mathbb{Q}_l/\mathbb{Z}_l)^n$ we have the s.e.s. of $\mathbb{Z}_l$-modules $$0 \to T \to A \to W\to 0.\tag{1}$$

Since $W$ is a discrete $G$-module, $H^i_c(G,W)=H^i(G,W)=0$ $(i>0)$ by the discrete case below and the l.e.s. of $(1)$ yields the surjection of $\mathbb{Z}_l$-modules

$$H^1_c(G,T) \twoheadrightarrow H^1_c(G,A).\tag{2}$$

[N,2.3.9] states:

Assume that the cohomology groups of $G$ with coefficients in finite $l$-primary modules are finite. Then $H^i_c(G,T)$ is a finitely generated $\mathbb{Z}_l$-module for all $i$.

By the discrete case below we can apply this theorem and find that $H^1_c(G,T)$ is a f.g.
$\mathbb{Z}_l$-module. Therefore, by $(2)$, the $\mathbb{Q}_l$-vector space $H^1_c(G,A)$ is f.g. as $\mathbb{Z}_l$-module what is only possible if $H^1_c(G,A)=0$. q.e.d.

**Discrete case:** Let $G$ be a pro-p group and $A$ a discrete $G$-module such that $A \to A, x \mapsto px$ is an automorphism. Then $H^i(G,A)=0$ for $i>0$. In particular, for each short exact sequence of discrete $G$-modules
$0 \to A \to B \to C\to 0$, the induced sequence $$0 \to A^G \to B^G \to C^G\to 0$$ is exact.

**Proof:** By the long exact cohomology sequence [RZ, 6.6.1] the latter follows from $H^1(G,A)=0$. Let $i>0$ and $x\in H^i(G,A)$. Since $H^i(G,A)= \varinjlim_U H^i(G/U,A^U )$ [RZ, 6.5.6], there is an open normal subgroup $U\le G$ and $y \in H^i(G/U,A^U)$ such that $x=\text{inf}(y)$. Since $G/U$ is a finite p-group, $y$ and hence $x$ is annulated by a power of p. But multiplication with p is an automorphism on $H^i(G,A)$. Thus $x=0$ and $H^i(G,A)=0$ follow. q.e.d.

[N]$\;\;$ Neukirch, et. al.: Cohomology of Number Fields.

[RZ] Ribes, Zalesskii: Profinite Groups, 2nd Edition.

notassuming the action was discrete. But now you say youareassuming it's continuous. But if $G$ is $p$-adic and we have a map $G\to GL(n,\mathbb{Q}_\ell)$ which is continuous then the image lands in a compact, which after conjugation can be $GL(n,\mathbb{Z}_\ell)$, and this has a finite index subgroup which is pro-$\ell$. This proves that any continuous action is discrete! – user30035 Feb 21 '13 at 9:17