My approach would be: first determine the time evolution of $\hat{x}(t)$ and $\hat{p}(t)$. For $\hat{x}$ you have $$ \frac{d}{dt}\hat{x}_H(t) = i[H_H,\hat{x}_H(t)] = \frac{i}{2m} [\hat{p}_H(t)^2,\hat{x}_H(t)] = \frac{\hat{p_H(t)}}{m} $$ and for $p$ you have (assuming $0\leq t \leq T$) $$ \frac{d}{dt}\hat{p}_H(t) = i[H_H(t),\hat{p}_H(t)] = -m\omega_0^2 \hat{x}_H(t) + F_0\sin(\Omega t) $$ These are coupled differential equations, which you can decouple by differentiating them once more with respect to time and performing a substitution. For instance,
$$ \frac{d^2}{dt^2} \hat{x}_H(t) = \frac{1}{m} \frac{d}{dt} \hat{p}_H = -\omega_0^2 \hat{x}_H(t)+\frac{F_0}{m}\sin(\Omega t) $$ where I substituted $\frac{d}{dt} \hat{p}_H(t)$ by its equation of motion found earlier. You can also get an equation like this for $\hat{p}_H(t)(t)$, which I leave for you..
Now, these equations can be solved using your favorite method, provided you give them suitable boundary conditions. 526
Note that you only need one boundary condition for $x$ and $p$ (which is $x_H(0)=\hat{x}_S$ and $p_H(0)=\hat{p}_S$ It will give you some expression for $\hat{x}_H(t)$ and $\hat{p}_H(t)$ in terms of $\hat{x}_S$ and $\hat{p}_S$. The Heisenberg Hamiltonian is then easily determined by substituting $\hat{x}_H(t)$ and $\hat{p}_H(t)$.
With that expression in hand you should be able to find $\langle H(t)\rangle$ (note that you should consider the cases where $t<0$ and $t>T$ separately).
EDIT: The proof regarding my statement below: In the <PERSON> picture the Hamiltonian is
$$\hat H_S=\frac{\hat{p}_S^2}{2m}+\frac{1}{2}m\omega_0^2\hat{x}_S^2-\hat{x}_SF_0\sin(\Omega t)$$
and the <PERSON> picture is given by $H_H = U^\dagger(t) H_S U(t)$. So if you take for instance the first term you get:
$$U^\dagger(t) \frac{\hat{p}_S^2}{2m}U(t) =\frac{1}{2m} (U^\dagger(t) \hat{p}_S U^\dagger(t))(U(t)\hat{p}_SU(t)) =\frac{1}{2m} \hat{p}_H(t)^2 $$
You can do the same for the other terms. In the end you just effectively replace $p_S\rightarrow p_H(t)$ and the same for $x$. 526