Using the notation of the OP, let $I(t)=\int_0^t X_s^2 ds$ where $X$ solves the above SDE. By Chebyshev's inequality, we have that
$$
P(I(t) > \alpha \mid X_0 = x) \le
\frac{E\left\{ I(t) \mid X_0 = x \right\}}{\alpha} \;.
$$
Fortunately, thanks to a Feynmann-Kac formula, the function
$$
u(t,x)=E\left\{ I(t) \mid X_0 = x \right\}
$$ appearing in the upper bound of this inequality is a local solution to an inhomogeneous, linear PDE:
$$
\begin{cases}
\partial_t u = L u + x^2 &\forall x, t\ge 0 \\
u(0,x) = 0 &\forall x
\end{cases} \tag{$\star$}
$$
where $L = - a x \partial_x + \frac{\sigma^2}{2} \partial_{xx}$ is the infinitesimal generator of the above SDE. Recall, that the linear operator $L$ has eigenvalues and eigenvectors:
$$
L e_k(x) = - k \cdot a \cdot e_k(x)
$$ where $e_k(x)$ is the $k$th Hermite polynomial and $k$ ranges over all natural numbers including zero. Expand the solution to $(\star)$ using these eigenvectors:
$$
u(t,x) = \sum_{k \ge 0} s_k(t) e_k(x)
$$
and similarly, expand the inhomogeneity:
$$
x^2 = (e_2(x) + 2 e_0(x) ) \frac{\sigma^2}{4 a}
$$
Substitute these expansions back into $(\star)$ and invoke orthogonality of these eigenvectors (in a weighted inner product space) to obtain the following system of ODEs for the spectral coefficients of $u(t,x)$:
$$
\begin{cases}
\dot s_0 = \frac{\sigma^2}{2 a} \;, & s_0(0) = 0 \;, \\
\dot s_1 = - a \cdot s_1 \;, &s_1(0) = 0 \;, \\
\dot s_2 = -2 \cdot a \cdot s_2 + \frac{\sigma^2}{4 a} \;, &s_2(0) = 0 \;, \\
\dot s_k = - k \cdot a \cdot s_k \;, &s_k(0) = 0 \;, \quad k \ge 3
\end{cases}
$$
The solutions to these ODES are all zero with the exception of:
$$
\begin{cases}
s_0(t) = \frac{\sigma^2}{2 a} t \\
s_2(t) = \sigma^2 \left( \frac{1 - e^{-2 a t}}{8 a^2} \right)
\end{cases}
$$ Thus, the solution to ($\star$) is:
$$
u(t,x) = \frac{\sigma^2}{2 a} t + \sigma^2 \left( \frac{1 - e^{-2 a t}}{8 a^2} \right) \left( \frac{4 a}{\sigma^2} x^2 - 2 \right) \;.
$$
To answer your question, take $x=0$ in this function to obtain:
$$
P(I(t) >\alpha \mid X_0 = 0) \le \frac{1}{\alpha}
\left(
\frac{\sigma^2}{2 a} t
- \sigma^2 \left( \frac{1 - e^{-2 a t}}{4 a^2} \right)
\right)
$$
You can adapt this technique to estimate all sorts of path-dependent expected values.