Multiplying both sides by $ (x - 2)(x + 2)(x^2 + 4), $ we get \[x^3 + 3x^2 - 12x + 36 = A(x + 2)(x^2 + 4) + B(x - 2)(x^2 + 4) + (Cx + D)(x - 2)(x + 2).\]Setting $ x = 2, $ we get $ 32A = 32, $ so $ A = 1 $. Setting $ x = -2, $ we get $ -32B = 64, $ so $ B = -2 $. Then \[x^3 + 3x^2 - 12x + 36 = (x + 2)(x^2 + 4) - 2(x - 2)(x^2 + 4) + (Cx + D)(x - 2)(x + 2).\]This simplifies to \[2x^3 - 3x^2 - 8x + 12 = (Cx + D)(x - 2)(x + 2),\]which factors as \[(2x - 3)(x - 2)(x + 2) = (Cx + D)(x - 2)(x + 2).\]Hence, $ C = 2 $ and $ D = -3, $ so $ (A,B,C,D) = \boxed{(1,-2,2,-3)} $.