Note that \[x^{12} + ax^8 + bx^4 + c = p(x^4).\]From the three zeroes, we have $ p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002) $. Then, we also have \[p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002).\]Let's do each factor case by case: First, $ x^4 - (2009 + 9002\pi i) = 0 $: Clearly, all the fourth roots are going to be nonreal. Second, $ x^4 - 2009 = 0 $: The real roots are $ \pm \sqrt [4]{2009} $, and there are two nonreal roots. Third, $ x^4 - 9002 = 0 $: The real roots are $ \pm \sqrt [4]{9002} $, and there are two nonreal roots. Thus, the answer is $ 4 + 2 + 2 = \boxed{8} $.