Solution:

By Vieta's formulas, $a + b + c = 0$ and $ab + bc + ca = -2011.$ Negating all three roots replaces $m$ by $-m,$ so we may assume two roots, say $a \ge b,$ are nonnegative. Substituting $c = -(a + b)$ into the second equation gives $ab - (a + b)^2 = -2011,$ that is, $$a^2 + ab + b^2 = 2011.$$

Multiplying by $4$ and completing the square, $(2a + b)^2 + 3b^2 = 8044.$ Since $a \ge b \ge 0,$ we have $3b^2 \le a^2 + ab + b^2 = 2011,$ so $0 \le b \le 25.$ Checking these values, $8044 - 3b^2$ is a perfect square only for $b = 10,$ where $8044 - 300 = 7744 = 88^2.$

Then $2a + b = 88$ gives $a = 39,$ and $c = -(a + b) = -49.$ (Indeed $39 \cdot 10 - 49^2 = -2011.$) Therefore $|a| + |b| + |c| = 39 + 10 + 49 = 98.$