Let me assume that the gamma distribution of each $\lambda_i$ has shape parameter $k_i>2$, so that $1/\lambda_i$ has a finite average and variance,
$${\mathbb E}(1/\lambda_i)=\frac{\Gamma(k_i-1)}{\theta_i\Gamma(k_i)}\equiv \mu_i,\;\;
{\rm Var}\,(1/\lambda_i)=\frac{1}{(k_i-2) (k_i-1)^2 \theta_i^2}\equiv\sigma_i^2$$
where $\theta_i$ is the scale parameter of the gamma distribution. The $n\times n$ matrix $R$ has elements
$$R_{ij}=\sum_{l=1}^m \frac{1}{\lambda_l} A^{\vphantom{\ast}}_{il}A^\ast_{jl}+\delta_{ij}/\gamma$$
$$\Rightarrow {\mathbb{E}}(R_{ij})=\sum_{l=1}^m \mu_l A^{\vphantom{\ast}}_{il}A^\ast_{jl}+\delta_{ij}/\gamma,$$
where $\ast$ denotes the complex conjugate.

Up to corrections of order $1/m$ we may replace the average of the inverse matrix by the inverse of the average matrix,
$$\mathbb{E}(R^{-1})=\bigl(\mathbb{E}(R)\bigr)^{-1}\bigl(1+{\cal O}(m^{-1})\bigr).$$
In particular, for $n=1$ one has
$$\mathbb{E}(R^{-1})=\frac{\gamma}{1+\gamma\sum_{l=1}^m \mu_l |A_{1l}|^2}\bigl(1+{\cal O}(m^{-1})\bigr).$$

An alternative approach, that avoids having to invert a matrix, is possible if $\gamma\ll 1$. A Taylor expansion in powers of $\gamma$ gives
$$R^{-1}=\gamma I_n+\gamma\sum_{p=1}^\infty(-\gamma)^p(A\Lambda^{-1}A^H)^p.$$
The expectation value to fourth order is
$${\mathbb{E}}\bigl((R^{-1})_{ij}\bigr)=\gamma \delta_{ij}-\gamma^2 \sum_{l=1}^m \mu_l A^{\vphantom{\ast}}_{il}A_{jl}^\ast$$
$$+\,\gamma^3\sum_{k,l,l'=1}^m (\mu_l \mu_{l'}+\delta_{ll'}\sigma^2_l)A_{il}^{\vphantom{\ast}}A_{kl}^\ast A_{kl'}^{\vphantom{\ast}}A_{jl'}^\ast+{\cal O}(\gamma^4).$$