I think that this follows from Bourbaki's *Éléments de Mathématique. Groupes et algèbres de Lie*, Chapitre VIII, §10, Theorem 1 (see below) applied to the adjoint representation.  Alas, I cannot provide the google books link because the book that Google Books claims to be this one, is actually *Algèbre commutative, Chapitres 5 à 7*!  (And the "Feedback" link does not allow me to point this out, since in their arrogance, Google does not even allow for the possibility of such an error!)

> *Théorème 1*. --- Soient $V$ un espace vectoriel de dimension finie, $\mathfrak{g}$ une sous-algèbre de Lie réductive dans $\mathfrak{gl}(V)$, $\mathfrak{q}$ une sous-algèbre de Lie de $\mathfrak{g}$ et $\Phi$ la forme bilinéaire $(x,y) \mapsto \mathrm{Tr}(xy)$ sur $\mathfrak{g} \times \mathfrak{g}$.  On suppose que l'orthogonal $\mathfrak{n}$ de $\mathfrak{q}$ par rapport à $\Phi$ est une sous-algèbre de Lie de $\mathfrak{g}$ composée d'endomorphismes nilpotents de $V$.  Alors, $\mathfrak{q}$ est une sous-algèbre parabolique de $\mathfrak{g}$.

And here's a possible translation:

> *Theorem 1*. --- Let $V$ be a finite-dimensional vector space, $\mathfrak{g}$ a reductive Lie subalgebra of $\mathfrak{gl}(V)$, $\mathfrak{q}$ a Lie subalgebra of $\mathfrak{g}$ and $\Phi$ the bilinear form $(x,y) \mapsto \mathrm{Tr}(xy)$ on $\mathfrak{g} \times \mathfrak{g}$.  If the orthogonal complement $\mathfrak{n}$ of $\mathfrak{q}$ relative to $\Phi$ is a Lie subalgebra of $\mathfrak{g}$ consisting of nilpotent endomorphisms of $V$, then $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}$.

**Edit**

As Fran points out in the comments below, my original translation was incorrect and had $\mathfrak{n}$ nilpotent instead of consisting of nilpotent endomorphisms.  <del>Happily, for the case of the adjoint representation, one has [Engel's theorem][1], which says that the the two notions agree.</del>


  [1]: http://en.wikipedia.org/wiki/Engel_theorem