More explicitly, theorem 2.7 of the paper in your notation says the following. Let $A$ be the adjacency matrix of $G_{n,p}$ multiplied by the factor $\gamma/q=\frac{1}{\sqrt{(1-p)pn}}$ where $p=q^2/n$ and $\gamma=(1-q^2/n)^{-1/2}$. Let $\mu_{n-1}$ be the second largest eigenvalue of $A$ (note that they index from the smallest to largest). Suppose that $q\geq n^\phi$ with $1/3<\phi\leq 1/2$. Then there exists $\delta>0$ such that for any $s$:
$F_1(s-n^{-\delta})-n^{-\delta}\leq PR[n^{2/3}(\mu_{n-1}-2)\leq s]\leq F_1(s+n^{-\delta})+n^{-\delta}$
Here $F_1(s)$ is the cumulative distribution function for the Tracy-Widom distribution for the orthogonal ensemble. The Wikipedia article I linked above has formulas for $F_1(s)$.
Because of the scale factor in front of $A$, their $\mu_{n-1}$ should be related to your $\lambda_2$ by the factor $\gamma/q=\frac{1}{\sqrt{(1-p)pn}}$. Thus after a bit of algebra, for large $n$:
$PR\left[\lambda_2\leq (s n^{-2/3}+2)\sqrt{(1-p)pn}\right]\approx F_1(s)$
Letting $x=(sn^{-2/3}+2)\sqrt{(1-p)pn}$, and noting that you want the probability that $\lambda_2$ is greater than $x$, if I haven't made any mistakes in transcription:
$PR\left[\lambda_2>x\right]\approx 1-F_1\left(n^{2/3}\left(\frac{x}{\sqrt{(1-p)pn}}-2\right)\right)$

