Setting $ x = -2 $ in the equation $ Q_1(x) (x + 2) - 13 = Q_3(x) (x + 2)(x^2 - 3x - 4) + R(x), $ we get \[R(-2) = -13.\]Setting $ x = 4 $ and $ x = -1 $ in the equation $ Q_2(x) (x^2 - 3x - 4) - 5x - 11 = Q_3(x) (x + 2)(x^2 - 3x - 4) + R(x), $ we get \[R(4) = -31 \quad \text{and} \quad R(-1) = -6.\]Since $ \deg R(x) = 2, $ we can let $ R(x) = ax^2 + bx + c $. Then \begin{align*} 4a - 2b + c &= -13, \\ 16a + 4b + c &= -31, \\ a - b + c &= -6.\end{align*}Subtracting these equations in pairs, we get \begin{align*} 12a + 6b &= -18, \\ 3a - b &= -7.\end{align*}Solving, we find $ a = -2 $ and $ b = 1, $ so $ c = -3 $. Hence, $ R(x) = \boxed{-2x^2 + x - 3} $.