Let $ \alpha $ be a root of $ x^3 - 3x^2 + 4x - 1 = 0, $ so $ \alpha^3 = 3 \alpha^2 - 4 \alpha + 1 $. Then \[\alpha^4 = 3 \alpha^3 - 4 \alpha^2 + \alpha = 3 (3 \alpha^2 - 4 \alpha + 1) - 4 \alpha^2 + \alpha = 5 \alpha^2 - 11 \alpha + 3.\]Hence, \begin{align*} \alpha^6 &= (3 \alpha^2 - 4 \alpha + 1)^2 \\ &= 9 \alpha^4 - 24 \alpha^3 + 22 \alpha^2 - 8 \alpha + 1 \\ &= 9 (5 \alpha^2 - 11 \alpha + 3) - 24 (3 \alpha^2 - 4 \alpha + 1) + 22 \alpha^2 - 8 \alpha + 1 \\ &= -5 \alpha^2 - 11 \alpha + 4, \end{align*}and \begin{align*} \alpha^9 &= \alpha^3 \cdot \alpha^6 \\ &= (3 \alpha^2 - 4 \alpha + 1)(-5 \alpha^2 - 11 \alpha + 4) \\ &= -15 \alpha^4 - 13 \alpha^3 + 51 \alpha^2 - 27 \alpha + 4 \\ &= -15 (5 \alpha^2 - 11 \alpha + 3) - 13 (3 \alpha^2 - 4 \alpha + 1) + 51 \alpha^2 - 27 \alpha + 4 \\ &= -63 \alpha^2 + 190 \alpha - 54.\end{align*}Then \begin{align*} \alpha^9 + p \alpha^6 + q \alpha^3 + r &= (-63 \alpha^2 + 190 \alpha - 54) + p (-5 \alpha^2 - 11 \alpha + 4) + q (3 \alpha^2 - 4 \alpha + 1) + r \\ &= (-5p + 3q - 63) \alpha^2 + (-11p - 4q + 190) \alpha + (4p + q + r - 54).\end{align*}We want this to reduce to 0, so we set \begin{align*} -5p + 3q &= 63, \\ 11p + 4q &= 190, \\ 4p + q + r &= 54.\end{align*}Solving, we find $ (p,q,r) = \boxed{(6,31,-1)} $. For these values, $ \alpha^9 + p \alpha^6 + q \alpha^3 + r $ reduces to 0 for any root $ \alpha $ of $ x^3 - 3x^2 + 4x - 1, $ so $ x^9 + px^6 + qx^3 + r $ will be divisible by $ x^3 - 3x^2 + 4x - 1 $.