I assume the matrices $X$ and $Y$ are independent. Since the trace commutes with the expectation value, and since the expectation value of the product of independent random variables is the product of expectation values, we have
$$
F(a)=\mathbb{E}\bigl[{\rm tr}\,\bigl({(aX{X^H} + I)^{ - 1}}Y{Y^H}\bigr)\bigr]={\rm tr}\bigl(\,\mathbb{E}[(aX{X^H} + I)^{ - 1}]\mathbb{E}[Y{Y^H}]\bigr).$$
The second factor is simply $K$ times the unit matrix, so
$$F(a)=K\,{\rm tr}\,\mathbb{E}[(aX{X^H} + I)^{ - 1}].$$
We can now again exchange trace and expectation value, to rewrite this as an integral over the eigenvalues $\mu_k$ of $XX^H$, with density $\rho(\mu)$,
$$F(a)=K\int \rho(\mu)(a\mu+1)^{-1}\,d\mu.$$
The density $\rho(\mu)$ is known, for large matrix size it is the Marcenko-Pastur distribution.