There is a long road to compute the expectation of the non covered area, and a shorter road. Both roads start with the random set $C$ of the points of the plane which are not covered by a meteorite and with the expression $ \mathrm E(|A\cap C|)=\mathrm E\int_A[x\in C]\,\mathrm dx=\int_A\mathrm P(x\in C)\mathrm dx=p\,|A|, $ with $p=\mathrm P(0\in C)$ since the invariance by translation of the meteorite process shows that $\mathrm P(x\in C)$ does not depend on $x$. Hence the task is to compute $p$.
The long road: Let $\mathcal M$ denote the random set of the centers of meteorites. Let $\mathcal R=(R(x))_{x\in\mathcal M}$ where, for every $x$ in $\mathcal M$, $R(x)$ denotes the radius of the meteorite centered at $x$. Then, $ \mathrm P(0\in C\mid \mathcal M,\mathcal R)=\prod_{x\in \mathcal M}[R(x)\lt\|x\|], $ hence, since conditionally on $\mathcal M$, $\mathcal R$ is a collection of i.i.d. random variables, $ \mathrm P(0\in C\mid \mathcal M)=\prod_{x\in \mathcal M}\mathrm P(R\lt\|x\|), $ where $R$ denotes any random variable distributed like the radii $R(x)$.
Now comes into play the hypothesis that $R$ is almost surely bounded, say by $r$. Then the product over $x$ in $\mathcal M$ can be restricted to a product over $\mathcal M\cap B$ where $B=\{x\mid\|x\|\leqslant r\}$. This new Poisson process has total intensity $\mu=\lambda\pi r^2$ and, conditionally on its size, its points are uniformly distributed in $B$. Thus, $ \mathrm P(0\in C)=\sum_{n\geqslant0}\mathrm e^{-\mu}\frac{\mu^n}{n!}\int_{B^n}\prod_{k=1}^n\mathrm P(R\lt\|x_k\|)\cdot\prod_{k=1}^n\frac{\mathrm dx_k}{\pi r^2}. $ The integral over $B^n$ is a product, hence $p=\mathrm e^{-\mu+\mu J/(\pi r^2)}$, with $ J=\int_{B}\mathrm P(R\lt\|x\|)\mathrm dx=\mathrm E\int_B[R\lt\|x\|]\mathrm dx=\mathrm E(\pi r^2-\pi R^2). $ Finally, $ p=\mathrm e^{-\lambda\pi\mathrm E(R^2)},\quad\text{hence}\quad\color{red}{\mathrm E(|A\cap C|)=\mathrm e^{-\lambda\pi\mathrm E(R^2)}\,|A|}. $ The shorter road: Consider the process of the centers of the meteorites covering the point $0$. This is a nonhomogenous Poisson process, obtained by a thinning of the original Poisson process, where one keeps a point $x$ with probability $\mathrm P(R\gt\|x\|)$. Thus the intensity at $x$ of the thinned Poisson process is $\lambda(x)=\lambda\mathrm P(R\gt\|x\|)$ and its total intensity is $ \nu=\int_{\mathbb R^n}\lambda(x)\mathrm dx=\lambda\int_{\mathbb R^n}\mathrm P(R\gt\|x\|)\mathrm dx=\lambda\mathrm E\int[R\gt\|x\|]\mathrm dx=\lambda\mathrm E(\pi R^2), $ To conclude, observe that $0$ is not covered if and only if the thinned Poisson process is empty, and that this happens with probability $\mathrm e^{-\nu}$.
Note: The argument above extends to any distribution of $R$. It shows that $\mathrm E(|A\cap C|)$ is always given by the formula above. Thus, if $R^2$ is integrable, $|A\cap C|$ has positive measure with positive probability, for every $A$ of positive measure, but, if $R^2$ is not integrable, $|C|=0$ almost surely.
The same applies to any dimension and to any random objects instead of disks. The condition to check in this extended setting is whether the random volume $V$ of these objects is integrable or not and the formula for any point $x$ of the space to be not covered becomes $ \color{purple}{\mathrm P(x\in C)=\mathrm e^{-\lambda\mathrm E(V)}}. $