For any interested latecomers who somehow discover this question in the future, I've found a lower-tech answer, bootstrapping from the low-tech answer to http://mathoverflow.net/questions/131270/is-a-lie-group-equivariantly-formal-under-conjugation-by-a-maximal-torus. I post this because I still don't understand the current answers. 

Anyway, once you believe that the conjugation action upon a compact Lie group $G$ of its maximal torus $T$ is equivariantly formal, it follows that the action of $G$ by conjugation is as well. This argument will actually work for any reasonably good space $M$ on which $G$ acts and the restricted $T$-action is equivariantly formal, because one has $$H_G(M) = H_T(M)^W = (H(M) \otimes H(BT))^W = H(M) \otimes H(BT)^W = H(M) \otimes H(BG)$$ as $H(BT)$-modules, where $W = N_G(T)/T$ is the Weyl group of $G$.

If, say, you don't buy that the action of $W$ on $H(M)$ is trivial, a longer proof goes like this.

\require{AMScd}

The homotopy quotient $M_G$ is a further quotient of $M_T$, and the projection $EG \times M \to EG$ then induces a commutative diagram

\begin{CD}
M @= M\\
@VVV @VVV\\
M_T @>>> M_G\\
@VVV @VVV\\
BT @>>> BG
\end{CD}

where the upper vertical maps are fiber inclusions.

The projection $BT = EG/T \to EG/G = BG$ induces an inclusion $H(BG) \cong H(BT)^W \hookrightarrow H(BT)$ in cohomology, and there are induced maps both in cohomology and on the Serre spectral sequences for the equivariant cohomologies, starting with this $E_2$ page:

\begin{CD}
H(M) @= H(M)\\
@AAA @AAA\\
H(M) \otimes H(BT) @<<< H(M) \otimes H(BG)\\
@AAA @AAA\\
H(BT) @<<< H(BG)
\end{CD}

Because the top and bottom horizontal maps are injective, so is the middle one, so the differentials for the spectral sequence converging to $H_G(M)$ are restrictions of those for $H_T(M)$. But the differentials for $H_T(M)$ are all zero, by equivariant formality, so the spectral sequence for $H_G(M)$ collapses as well.