Actually, no matter how large $A$ 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, when $A\ge0$, there is one real root $r>0$ and two complex conjugate roots $z$ and $\bar z$.  These roots satisfy $z\bar z r = 1$ and $z\bar z + r(z+\bar z) = 0$, so we have $z+\bar z  = -1/r^2 < 0$, so the real part of $z$ is negative.  Hence, there is a stable manifold of dimension 2 (and an unstable manifold of dimension 1), say $M^2\subset\mathbb{R}^3$ that passes through $(0,0,0)$ and is such that any initial condition starting on $M$ will spiral in to $(0,0,0)$.  Thus, such solutions will stay bounded for all time, in fact, they'll decay to zero.