You're taking cohomology of classifying spaces, aren't you? I think the answer is positive, even if $A$ is not finite. The argument I have in mind is as follows. The short exact sequence
$$G_c\hookrightarrow G\twoheadrightarrow G/G_c=\pi_0G$$
induces a fibration after taking classifying spaces
$$BG_c\hookrightarrow BG\twoheadrightarrow B\pi_0G.$$
This, in turn, gives rise to a Serre spectral sequence
$$E_2^{p,q}=H^p(B\pi_0G,H^q(BG_c,A))\Longrightarrow H^{p+q}(BG,A).$$
The spectral sequence differential $d_r$ has bidegree $(r,1-r)$ and the cohomology groups $H^q(BG_c,A)$ are regarded here as $\pi_0G$-modules.
The morphism whose infectivity you want to prove decomposes as
$$E_2^{3,0}=H^3(B\pi_0G,A)\twoheadrightarrow E_\infty^{3,0}\hookrightarrow H^3(BG,A).$$
Hence, we just have to prove that $E_2^{3,0}=E_\infty^{3,0}$. In principle, $E_\infty^{3,0}$ is computed from $E_2^{3,0}$ in two steps, taking two quotients. More precisely, we have three exact sequences
$$E_2^{1,1}\stackrel{d_2}{\longrightarrow}E_2^{3,0}\twoheadrightarrow E_3^{3,0}$$
$$E_3^{0,2}\stackrel{d_3}{\longrightarrow}E_3^{3,0}\twoheadrightarrow E_4^{3,0}=E_\infty^{3,0}$$
$$E_3^{0,2}\hookrightarrow E_2^{0,2}\stackrel{d_2}{\longrightarrow}E_3^{2,1}$$
Therefore, if we prove that $E_2^{1,1}=0=E_2^{0,2}$ we will be done.
The first one is $$E_2^{1,1}=H^1(B\pi_0G,H^1(BG_c,A)).$$ This one vanishes because $G_c$ is connected hence $BG_c$ is simply connected so $H^1(BG_c,A)=0$.
The second one is $$E_2^{0,2}=H^0(B\pi_0G,H^2(BG_c,A)).$$
Since $BG_c$ is simply connected, $H^2(BG_c,A)=\hom(H_2(BG_c),A)$. Moreover, $H_2(BG_c)=\pi_2BG_c=0$. This concludes the argument.