You're taking cohomology of classifying spaces, aren't you? I thought first that the answer was positive, but there was a mistake in my argument. I can recycle the argument to show that there is an exact sequence

$$H^2(B\pi_0G,A)\hookrightarrow H^2(BG,A)\rightarrow H^2(BG_c,A)^{\pi_0G}\rightarrow H^3(B\pi_0G,A)\rightarrow H^3(BG,A)$$

from which one can extract various necessary and/or sufficient condition for your map (the last one in the sequence) to be injective. Here all maps, except for the third one, are induced by morphisms in the short exact sequence below.

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 injectivity you want to prove decomposes as

$$E_2^{3,0}=H^3(B\pi_0G,A)\twoheadrightarrow E_\infty^{3,0}\hookrightarrow H^3(BG,A).$$

We have exact sequences

$$E_2^{1,1}\stackrel{d_2}{\longrightarrow}E_2^{3,0}\twoheadrightarrow E_3^{3,0}$$

$$E_3^{0,2}\hookrightarrow E_2^{0,2}\stackrel{d_2}{\longrightarrow}E_2^{2,1}$$

$$E_2^{0,1}\rightarrow E_2^{2,0}\twoheadrightarrow E_3^{2,0}=E_\infty^{2,0}$$

$$E_\infty^{0,2}=E_4^{0,2}\hookrightarrow E_3^{0,2}\stackrel{d_3}{\longrightarrow}E_3^{3,0}\twoheadrightarrow E_4^{3,0}=E_\infty^{3,0}$$

The group
$$E_2^{n,1}=H^n(B\pi_0G,H^1(BG_c,A))=0$$
vanishes because $G_c$ is connected hence $BG_c$ is simply connected so $H^1(BG_c,A)=0$. Therefore

$$E_3^{3,0}=E_2^{3,0},\quad E_3^{0,2}=E_2^{0,2},\quad E_\infty^{2,0}=E_2^{2,0}.$$

Moreover, we have a short exact sequence

$$E_\infty^{2,0}\hookrightarrow H^2(BG,A)\twoheadrightarrow E_\infty^{0,2}.$$

Splicing some of the previous exact sequences we get an exact sequence

$$E_2^{2,0}\hookrightarrow H^2(BG,A)\rightarrow E_2^{0,2}\rightarrow E_2^{3,0}\rightarrow H^3(BG,A).$$

This is the first exact sequence, above. We just have to notice that 

$$E_2^{0,2}=H^0(B\pi_0G,H^2(BG_c,A))=H^2(BG_c,A)^{\pi_0G}.$$