How do I add AWGN to an I and Q representation of a signal?
Yes, you can add AWGN of variance $\sigma^2$ separately to each of the two terms, because the sum of two Gaussians is also a Gaussian and their variances add up. This will have the same effect as adding an AWGN of variance $2\sigma^2$ to the original signal. Here's some more explanation if you're interested.

An analytic signal $x(t)=a(t)\sin\left(2\pi f t + \varphi(t)\right)$ can be written in its in-phase and quadrature components as

$$x(t)=I(t)\sin(2\pi ft) + Q(t)\cos(2\pi ft)$$

where $I(t)=a(t)\cos(\varphi(t))$ and $Q(t)=a(t)\sin(\varphi(t))$. If you wish to add AWGN to your original signal as $x(t)+u(t)$, where $u(t)\sim\mathcal{N}(\mu,\sigma^2)$, then you can add AWGN to each of the terms as 

$$y_1(t)=\left[I(t)\sin(2\pi ft) + v(t)\right] + \left[Q(t)\cos(2\pi ft) + w(t)\right]$$

where $v(t), w(t)\sim\mathcal{N}(\mu/2,\sigma^2/2)$

Also note that because the in-phase and quadrature terms are additive, the AWGN can also be simply added to either of the two terms in the $IQ$ representation of $x(t)$ above. In otherwords,

$$y_2=I(t)\sin(2\pi ft) + \left[Q(t)\cos(2\pi ft) + u(t)\right]$$
$$y_3=\left[I(t)\sin(2\pi ft) +u(t)\right]+ Q(t)\cos(2\pi ft)$$

are statistically equivalent to $y_1$, although I prefer using $y_1$ because I don't have to keep track of which component has noise added to it.