EDIT: My reasoning below seems to be wrong. The process as you write it tends to infinity if $a$ is big enough and positive and if $\lambda_0$ is positive.
I would not call this process non-meanreverting OU. It is just an Ito process of a simple form. If we remove the stochastic part then we get
$$
d\lambda_t = a \lambda_t dt
$$
with the solution (if $\lambda_0>0$) $\lambda_t = \lambda_0 \exp(a t)$ which for $a>0$ just grows exponentially.
If look at the whole thing then we add a stochastic disturbance at each time step of size $\sigma dB_t$.
Thinking about it this way I think that the process above does not have too much in common with an OU-process.
I delete my previous answer.
Concerning the estimation: If you have a process that you have observed on a time grid with width $ \Delta t$ then a discretization of your SDE could look like this:
$$
\lambda(t + \Delta t) - \lambda(t) = \theta \lambda(t) \Delta t + \sigma \sqrt{\Delta t} \epsilon_i
$$
where $\epsilon_i$ is standard normal.
Thus a regression of $\lambda(t + \Delta t) - \lambda(t)$ on $\lambda(t)$ gives you $\theta \Delta t$. The volatility of the residuals gives you an estimate of $\sigma \sqrt{\Delta t}$. Dividing these quantities by the grid width (resp its square-root) gives you the parameters.
Look at a similar question here.