I don't know if this exactly works for you, but will give the relaxed version a shot:
Preliminaries:
Correlation matrix can be seen as the covariance matrix of the standardized random variables $X_{i}/\sigma (X_{i})$. And, any correlation matrix can be converted to a covariance matrix as:
$$Q={\text{corr}}({X} )=\left({\text{diag}}(\Sigma )\right)^{-{\frac {1}{2}}}\,\Sigma \,\left({\text{diag}}(\Sigma )\right)^{-{\frac {1}{2}}}$$
If standard deviations $\{\sigma_i\}$ are known, then using $D = \sqrt{diag(\Sigma)}$:
$$
\Sigma = DQD.
$$
Formulation:
Using the definition above we can equivalently write $Q = D^{-1}\Sigma D^{-1}$. If we use this definition of $Q$ and set $B=D^{-1}$, then:
$$
\begin{align}
x^T Q x &= x^T B^T \Sigma B x \\
&= x^T B^T V^T \Lambda V B x \\
&= x^T B^T V^T \sqrt{\Lambda}^T\sqrt{\Lambda} V B x \\
\end{align}
$$
$\Sigma=V^T \Lambda V$ follows from the eigen-decomposition and nicely exists for the covariance matrix. Then, substituting $M = B^T V^T \sqrt{\Lambda}^T$, one can show that:
$$
x^T Q x = x^T M^T M x = \lVert M x \rVert^2.
$$
The problem then turns to :
$$
\arg\min_x \lVert M x \rVert^2 + \lambda (\|x\|_0-m)
$$
which is generally tackled via the $L_1$ variant of the form:
$$
\arg\min_x \lVert M x \rVert^2 + \lambda \|x\|_1.
$$
Now instead of thinking about $x$ as a binary vector, we could relax this constraint and seek for the sparsest solution under real $x$. This is a non-quadratic $L_1$-regularized Least Squares problem and one can solve it by standard techniques, such as (I guess) Lasso. Maybe it is then possible to look at the solution and determine $m$. This gives local optimality but should work well in practice. For hard constraints Lagrange multipliers can be used.
I might have to double check the formulation. As I don't have much time now, I'll do it later, but the method above should give you the idea.