First, the condition that the kernel of $\alpha$ be integrable is not just that the matrix $A$ be symmetric although, of course, that is sufficient. For example, when the dimension $n$ is equal to $2$, the $1$-form is always integrable, even if $A$ is skew-symmetric.
Second, the condition that the kernel of $\alpha$ be bracket-generating is just that $\alpha\wedge\mathrm{d}\alpha$ should be nonvanishing. This will always be true (except where $\alpha$ vanishes) if $(\mathrm{d}\alpha)^2$ is nonzero. Writing $A = B + C$ where $B$ is symmetric and $C$ is skew-symmetric, and correspondingly, writing $\alpha = \beta + \gamma$, then $\beta = \mathrm{d}b$ for some quadratic function $b$ on $\mathbb{R}^n$ and $\mathrm{d}\alpha = \mathrm{d}\gamma$.
Thus, if the rank of $C$ is greater than $2$, then $\alpha\wedge\mathrm{d}\alpha$ will be nonzero on the open dense set (the complement of a linear subspace, the kernel of $A$) where $\alpha$ is nonvanishing, and hence the kernel of $\alpha$ will be bracket-generating on this open set in this case.
If $C=0$, then $\mathrm{d}\alpha = 0$, and the kernel of $\alpha$ is integrable.
If the rank of $C$ is equal to $2$, then one can choose linear coordinates $y^1,\ldots,y^n$ on $\mathbb{R}^n$ such that $\gamma = \tfrac12(y^1\,\mathrm{d}y^2-y^2\,\mathrm{d}y^1)$, and then $\gamma\wedge\mathrm{d}\gamma=0$, so that
$$\alpha\wedge\mathrm{d}\alpha
= \beta\wedge\mathrm{d}\gamma
= \mathrm{d}b\wedge\mathrm{d}\gamma
= \mathrm{d}b\wedge \mathrm{d}y^1\wedge\mathrm{d}y^2.
$$
This latter expression vanishes if and only if $b$ is a quadratic form in the variables $y^1$ and $y^2$. In particular, $\alpha$ is then a $1$-form expressed in terms of $y^1$ and $y^2$ alone. Thus, we see that $\alpha\wedge\mathrm{d}\alpha = 0$ with $\mathrm{d}\alpha\not=0$ if and only if there is a linear map $f:\mathbb{R}^n\to\mathbb{R}^2$ such that $\alpha=f^*\bar\alpha$,
where $\bar\alpha$ is a non-closed $1$-form on $\mathbb{R}^2$ with linear coefficients.
One might be interested in the case in which the kernel of $\alpha$ is a contact plane field, which is a stronger condition.
Now, in order for the kernel of $\alpha$ to be contact (away from the origin, where, of course, the kernel of $\alpha$ is everything), we must have that the dimension $n$ be odd, say, $n=2m+1$ and that $\alpha\wedge(\mathrm{d}\alpha)^m$ be nonzero (away from the origin). Again, writing $A = B + C$ where $B$ is symmetric and $C$ is skew-symmetric, and, correspondingly, $\alpha = \beta + \gamma$, then $\beta = \mathrm{d}b$ for some quadratic function $b$ on $\mathbb{R}^n$ and $\mathrm{d}\alpha = \mathrm{d}\gamma$.
Now, it is not hard to show that $\gamma\wedge(\mathrm{d}\gamma)^m$ vanishes identically, so that
$$
\alpha\wedge(\mathrm{d}\alpha)^m = \beta\wedge(\mathrm{d}\gamma)^m.
$$
Thus, the contact condition implies that $(\mathrm{d}\gamma)^m$ be nowhere vanishing. It follows that, by a linear change of coordinates, we can find coordinates $y^0,y^1,y^2,\ldots,y^{2m}$ such that
$$
\gamma =\tfrac12\bigl( y^1\,\mathrm{d}y^2-y^2\,\mathrm{d}y^1+y^3\,\mathrm{d}y^4-y^4\,\mathrm{d}y^3 + \cdots + y^{2m-1}\,\mathrm{d}y^{2m}-y^{2m}\,\mathrm{d}y^{2m-1}\bigr).
$$
When we write $b$ in these coordinates, it has the form
$$
b = \tfrac12\bigl(b_{0}(y^0)^2 + 2b_{i}y^0y^i + b_{ij}\,y^iy^j\bigr),
$$
and so
$$
\alpha \wedge (\mathrm{d}\alpha)^m = m!\, (b_0y^0+b_iy^i) \,\mathrm{d}y^0\wedge\mathrm{d}y^1\wedge\cdots\wedge\mathrm{d}y^{2m}.
$$
Thus, we see that, in order to have the kernel be contact (at least away from a hyperplane), we need that $C$ have rank $2m = n-1$ (the maximum that it could possibly have) and that null space of the quadratic form defined by $B$ should not contain the ($1$-dimensional) kernel of $C$. This is necessary and sufficient for contact (away from the obvious hyperplane).
Finally, let me remark that, in order to get a sub-Riemannian structure, you also need a positive definite quadratic form defined on the kernel of $\alpha$. Now, it's not immediately clear which one you want to take. In the generic case, you could simply use $B$ to define a quadratic form $g_B = b_{ij}\,\mathrm{d}x^i\mathrm{d}x^j$ and restrict it to the kernel of $\alpha$. Assuming that this is positive definite on the kernel of $\alpha$ (which, of course, it would be if $B$ were positive definite), this would work. Is this the sub-Riemannian structure you have in mind?
