Upper bound on Independence Number of Random Regular Graph with degree $\Theta(\sqrt{|V|} \log^2 |V|)$
I don't think the Bollobás paper asserts the bound on the independence number; rather, it seems to me that it asserts that for any given maximum degree $\Delta$ and lower bound on the girth $g$, there exists graphs of degree at most $\Delta$ and girth at least $g$ with independence ratio at most $2\log \Delta/\Delta$. In contrast, as you mention, the Frieze and Łuczak paper seems to give concentration of random $\Delta$-regular graphs that holds for $\Delta=o(n^{1/3})$, so that the desired upper bound holds with probability $1-o(1)$. This result is thus incomparable to the Bollobás paper, I think.
For your specific application, I think one can give an upper bound of roughly $\alpha(G)\leq O(n/\sqrt{\Delta})=\tilde{O}(n^{3/4})$ that holds with probability $1-o(1)$ (in fact, at least inverse polynomial error probability). I have no idea if this bound is closer to the truth for this regime, and this is of course much weaker than the bound that you want (note that no bound better than $\tilde{O}(n^{1/2})$ is possible anyway, for any $\Delta$-regular graph with $\Delta=\tilde{O}(n^{1/2})$-here, I'm using the tilde notation to mean up to logarithmic factors, in either direction).
Here's two ways that I hope work to get this bound: one is to note that the Frieze and Łuczak paper upper bound does not seem to rely on the constant degree restriction. In particular, just mimicking their proof of the upper bound (which is by far the easier direction in their concentration result), it seems to me like everything should go through to show that the probability of having an independent set of size greater than $\tilde{O}(n^{3/4})$ is $1-o(1)$. The reason their bound degrades this much when the degree is not constant is because it incurs a $e^{\Delta^2}$ loss when moving between simpler random graph models.
Another way to get it is to use some recent high-powered results that show that if $\Delta\to \infty$ with $n$ but $\Delta=o(n^{1-\epsilon})$ for some fixed $\epsilon&gt;0$, then the spectral radius of the adjacency matrix (excluding the trivial eigenvalue $\Delta$) is $O(\sqrt{\Delta})$ with probability $1-o(1)$ (where the constant depends on $\epsilon$, but not $\Delta,n$). If $\lambda=\lambda(G):=\max\{\vert \lambda_2(G)\vert,\vert \lambda_n(G)\vert\}$ is the second largest eigenvalue of the adjacency matrix of a graph $G$, Hoffman's bound asserts that for a $\Delta$-regular graph on $n$ vertices:
\begin{equation}
\alpha(G)\leq \frac{\lambda}{\Delta-\lambda}n.
\end{equation}
For $\Delta=\tilde{O}(n^{1/2})$ and using the estimate above gives with probability $1-o(1)$ that $\alpha(G)\leq O(n/\sqrt{\Delta}) = \tilde{O}(n^{3/4})$.