The line from your textbook you're puzzling over is in fact quite a bit easier: in a stationary state nothing changes apart from phases that cancel out in any expectation value, and therefore $\frac{d}{dt}\left[\textrm{anything} \right]=0.$
You're right to point out that $xp$ is not a hermitian operator, but that does not mean that the expectation value $\langle xp \rangle$ is meaningless. Specifically, take the uncertainty relation $xp-px=i\hbar$ and substract $xp$ twice: you get $$xp+px=-i\hbar+2xp,$$ which you can rearrange into $$\langle xp\rangle = \left\langle \frac{xp+px}{2}\right\rangle+\frac{i\hbar}{2}.$$
The expectation value is now of the hemitian operator $\frac{1}{2}(xp+px)$, and you can see that your original expectation value has a trivial imaginary part.
If it's a physical interpretation for this quantity you're after, try this question.
In position-space (that is, when your functions are functions of x), the function $\int|\Psi|^2$ gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, $|\Psi|^2$, is the weighting function (that's not exactly what it is, but it's a useful analogy). Similarly, you can find the expectation value for any measurable quantity. In this space, the difference between the two is that the expectation value is a number that represents the expected average position of the particle over many measurements whereas the probability is a number that gives you the probability for finding the particle within the limits of integration.
However, you can use any different basis. For example, you could choose momentum-space, $\left|\Psi\right>$ is $\Psi(p)$ (quantum physicists please don't kill me for that affront to notation). In momentum space, the integral $\int|\Psi|^2$ is now the probability of the particle having a given range of momenta. However, the expectation value of x is still the average measurement of x. What, you ask, is the point? The expectation value is a number that can be found in any basis that represents the "on-average" value of a measurement. The probability found by $\int|\Psi|^2$ is the probability that a particle will be found existing within a specified range of values for the basis you are using.
$\int_{x_1}^{x_2}|\Psi|^2dx$ is "there is #% chance that the particle will be found between $x_1$ and $x_2$"
$\left<\Psi\right|x\left|\Psi\right>$ is "the expected average position of the particle over a large number of sample measurements is at $x$=#"
$|\Psi|^2(x)$ is a function "the probability per unit length of finding the particle at this position is #%"
Best Answer
In the same way that the expectation value of the position operator is the average position you'd get if you measured a bunch of identically-prepared states, the expectation value of the Hamiltonian operator is the average value of the Hamiltonian that you'd get if you measured a bunch of identically-prepared states. In most of the elementary situations you'll be looking at,* the value of the Hamiltonian is equivalent to the total energy, so the expectation value of the Hamiltonian is the average value of the energy that you'd get if you measured a bunch of identically-prepared states.
*The question of when the Hamiltonian is equivalent to the total energy is a complicated one, and depends, in part, on what you define "the total energy" to be in the first place, but until you get into Hamiltonians involving the electromagnetic field, you can usually take the two to be equivalent.