For a regular Poisson process, we know that the inter-arrival distribution, $T$ is exponential with rate $\lambda$ and the number of events in any interval, $N(t)$ is Poisson with mean $\lambda t$ and variance $\lambda t$ as well. Let's say we observe the process for a certain amount of time, $t$. The unbiased estimator of $\lambda$ will become:
$$\hat{\lambda} = \frac{N(t)}{t}$$
The expected value of this estimator is: $E(\hat{\lambda}) = \frac{E(N(t))}{t} = \lambda$ as expected.
And the variance of this estimator will be:
$$V(\hat{\lambda}) = \frac{V(N(t))}{t^2} = \frac{\lambda}{t}$$ and this is encouraging since for a large timeframe of observation, the variance in this estimator will go to $0$.
Now, I want to consider a Mixed Poisson process, described in section 5.4.3 of Ross' Introduction to Probability Models. It is a regular Poisson process when conditioned on some distribution $L(\lambda)$ of the rate, $\lambda$. Now, we're still interested in calculating the average hazard rate of this process. It's clear by definition (conditional on $L$, we get the regular Poisson process):
$$E(N(t)|L)=Lt$$ $$V(N(t)|L)=Lt$$
Using the law of total expectation:
$$E(N(t)) = t E(L)$$ Using the law of total variance:
$$V(N(t))=E(V(N(t)|L))+V(E(N(t)|L))$$ $$=E(Lt)+V(Lt)$$ $$=tE(L)+t^2V(L)$$
This means that if we observe this process for a large period of time, $t$, we can estimate the average hazard rate:
$$\hat{\lambda} = \frac{N(t)}{t} $$ $$=>E(\hat{\lambda})= E(L)$$
And the variance of this estimator becomes:
$$V(\hat{\lambda}) = \frac{V(N(t))}{t^2} = \frac{E(L)}{t}+V(L)$$
Now, this is worrying since no matter how large $t$ becomes, the variance of the estimator, $\hat{\lambda}$ never goes to zero. Is there a different way then, to estimate $E(\hat{\lambda}) = E(L)$ where we can be sure that given a large enough time-frame of observation, we will get an unbiased estimator with variance tending to zero?
I think this is a conceptual problem rather than a question of finding a better estimator. You can't reduce the variance of $\hat\lambda$ below $V(L)$, since by definition the rate depends on $L$ and without any information about $L$ you don't know the exact rate.
I get the impression that what you're actually trying to do is perhaps to estimate the rate of a given instance of the process with arbitrary precision. You can do that using your estimator. Its variance doesn't reflect uncertainty about the rate in any given instance of the process, but uncertainty about which rate the process will have.
For instance, in Example $5.30$ on p. $353$ of the text you linked to, you can determine the claim rate of any given customer to arbitrary precision by observing them for long enough. But the variance you calculated is the variance over all customers.