Actually, no matter what $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded.  Here is why:  First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x_0$, $\dot x = x_1$, and $\ddot x = x_2$.  Then the equation becomes the first order system
$$
\begin{aligned}
\dot x_0 &= x_1\,,\\
\dot x_1 &= x_2\,,\\
\dot x_2 &= -x_0 + {x_1}^2 - A\,x_2\,.
\end{aligned}
$$
This vector field (i.e., ODE) has one singular point, $(x_0,x_1,x_2) = (0,0,0)$,
and its linearization at this point has the matrix
$$
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
-1 & 0 & -A
\end{pmatrix}
$$
The eigenvalues of this matrix are the roots of  $\lambda^3+A\lambda^2 + 1 = 0$, and there is always at least one negative real root.  Hence the stable manifold of $(0,0,0)$ has dimension at least $1$, so there will always be a nonzero solution that decays to zero (in infinite time).