Energy conservation $\iff \frac{dE}{dt} = 0\ $? If I'm asked to prove that a system is/ isn't conservative and compare it to whether or not the Hamiltonian is conserved, does that mean I need to compute the time derivative of energy $(T+U)$?  Doing so in my current problem will just result in a long expression in $\dfrac{dq}{dt}$'s and $\dfrac{dp_q}{dt}$'s, and I wouldn't be certain whether it'd have to be $0$ or not.
Is there some other way of telling whether the system is conservative?
 A: You need to use the equations of motion in the expression for $dE/dt$.
Lets take a simple example. For the Hamiltonian
$$\mathcal{H} = \frac{1}{2}\dot{x}^2 + V(x)$$ 
(which is equivalent to the Lagrangian $\mathcal{L} = \frac{1}{2}\dot{x}^2 - V(x)$) we have that the time-derivative of the energy is
$$\frac{d\mathcal{H}}{dt} = \dot{x}\ddot{x} + V'(x)\dot{x}$$ 
This can be any value, but the particle this system describes follows only certain paths given by the equations of motion. To show that $d\mathcal{H}/dt$ is indeed zero you need to first derive the equations of motion for the system and use this in the relation above. The Hamiltonian (or Lagrangian) equations gives
$$\ddot{x} + V'(x) = 0$$
which gives the desired result
$$\frac{d\mathcal{H}}{dt} = \dot{x}[\ddot{x} + V'(x)] = 0$$ 
A: If the expression for the energy $E(t)$ is analytic, and this energy is not conserved, then for all but a discrete set of $t$ it will be different from $E(0)$.  Pick some random time $t$ and evaluate $E(0)$ and $E(t)$ sufficiently accurately, and you will find that they are different.
On the other hand, if the energy is conserved, it is conceivable that this is due to an identity that you don't recognize: in a sufficiently rich system of
functions there is no decision procedure for identities.  But in practice I think it'll be pretty rare: in "most" conservative systems you're likely to encounter, it won't be hard to simplify $dE/dt$ to $0$.
