Do the eigenvectors of a random orthogonal matrix have Haar measure? For orthogonal $Q$ with Haar measure, does the group of unitary matrices $U$ which diagonalize $Q=U\Lambda U^H$ have Haar measure? I'd be happy to know any answer, even if it's only true for certain cases (special orthogonal, even, odd, etc.). 
---Edit---
I think Pierre presented a nice argument that $U$ should fail with respect to Haar measure on the unitary group. I have a follow-up observation about the character of $U$ based on the following decomposition of $Q$ which exhibits Haar measure on $SO(2n)$.
$$
Q=Y \, \text{diag}(R_{\theta_1},\dots,R_{\theta_1}) Y^T
$$
In this expression, I believe $Y$ is uniform on $SO(2n)$ as well. The $R_{\theta_i}$'s are the $2\times2$ rotations associated with the conjugate pairs of eigenvalues of $Q$.  
$$
\text{diag}(R_{\theta_1},\dots,R_{\theta_n})  = \frac{1}{2}\text{diag}\left(\begin{bmatrix}
1 & 1\\
i & -i
\end{bmatrix}
\begin{bmatrix}
e^{i\theta} &\\
 & e^{-i\theta}
\end{bmatrix}
\begin{bmatrix}
1 & -i\\
1 &  i
\end{bmatrix},\dots \right)
$$
This means that $U$ can be written as
$$
U = \frac{1}{\sqrt{2}}Y\text{diag}\left(\begin{bmatrix}
1 & 1\\
i & -i
\end{bmatrix},\dots \right)
$$
I think that this implies that the dimension of $U$ is the same as the dimension  of $Y$ which is uniform on $SO(2n)$ and therefore $n(2n-1)$. If $U$ were Haar-uniform the unitary group on  it's dimension would be $4n^2$.
I apologize for any abuse in notation/terminology (feel free to correct me!); I'm not really a mathematician.
 A: This is by no means a complete solution,
but I can offer what I think is at least a plausibility argument for why the claim should not be true.
Let $Q$ be an orthogonal matrix,
and let $Q=UDU^*$ be its unitary diagonalization ($*$ denotes the Hermitian transpose).
Then,
the fact that $Q^TQ=I$ implies that
$$\bar{U}DU^TUDU^*=I,$$
(where $\bar{U}$ denotes the entry-wise complex conjugate, or the transpose of the conjugate-transpose)
and thus
$$DU^TUD=U^TU=D^{-1}U^TUD^{-1}.\tag{$\ast$}$$
Now,
equation $(\ast)$ will of course hold if $D=D^{-1}$ (i.e., all the eigenvalues of $Q$ are either $1$ or $-1$), or of $U^TU=I$ (i.e., all the entries in $U$ are real-valued). However, I think one could show without too much difficulty that $D=D^{-1}$ with probability zero (at least as $n\to\infty$) with respect to the Haar measure on the orthogonal group (indeed, it is well-known that the empirical eigenvalue distribution of Haar distributed random matrices on the orthogonal group converges almost-surely to the uniform measure on the unit disc in $\mathbb C$), and that $U^TU=I$ with measure zero according to the Haar measure on the unitary group.
Then,
all that would remain to show is that for some generic diagonal matrix $D$ whose diagonal elements are more or less distributed uniformly on the unit disk,
$(\ast)$ fails with positive probability with respect to the Haar measure on the unitary group.
