In $R^5$ with coordinates $(p_1,q_1,p_2,q_2,t)$, $da=dp_1\wedge dq_1$. Ignore the orientation. Let $S=\{t\in[0,1],p_1^2+q_1^2=t^2,p_2^2+q_2^2=t^2\}$
$\int_Sda\wedge dt$. So $\partial S=\{t=1,p_1^2+q_1^2=p_2^2+q_2^2=1\}$.
=$\int_{\partial S} tdp_1\wedge dq_1$ by $d(tdp_1\wedge dq_1)=da\wedge dt$
=$0$ as pulling back $dt$ is zero by $(p_1,q_1,p_2,q_2,t)\to (r cos(t_1),rsin(t_1),rcos(t_2),rsin(t_2),1)$.
The other way is to see that $C=\{t\in[0,1], p_1^2+q_1^2=p_2^2+q_2^2\leq t^2\}$. And $\partial C=S\cup S'$ where $S'=\{t=1,p_1^2+q_1^2=p_2^2+q_2^2\leq 1\}$
So $\int_{S\cup S'}da\wedge dt$=0 by stokes theorem. Now $\int_{S'}da\wedge dt=0$ by the same reason as above.
Did I make any mistake here as I though I should get a non zero number? However, I feel this integral is obtain something two dimensional volume in 3 dimensional space. And thus I expect it to be zero.
The answer is $0$, but the rationale you gave is wrong. $t=1$ on the boundary, but the integral of $dp_1\wedge dq_1$ over the unit circle (not disk) in $p_1q_1$-space is clearly $0$. If you'd taken $da=dp_1\wedge dp_2$, things would come out the same, but for different reasons.