Minimum set cover with incompatible sets
Your problem can be stated as a minimum weight maximal independent set problem.
Construction:
Construct a bipartite graph $G = (L,R,E)$, where the right partition $R$ corresponds to $\mathcal{R}$ and the left partition $L$ corresponds to $X$. An edge $(u,v) \in E$ iff an element $u$ is contained in the set $v$.
Let $G'$ be the updated graph obtained by adding edges within $R$ based on the incompatibility of vertices in $R$.
Furthermore, assign a weight of $\infty$ on each vertex of $L$ and weight $1$ on each vertex of $R$. Also, attach a new pendant vertex of weight $0$ to each vertex in $R$. Let $G_{w}$ be this new graph.
Now, your problem can be stated as finding minimum weight maximal independent set problem (MMIS) on $G_{w}$. It is a standard problem discussed here. Your problem has a feasible solution if and only if MMIS has a finite value. Also, your problem has a solution of size $k$ if and only if MMIS has a value $k$.
Correctness:
Suppose, a feasible instance of the independent set cover problem is $S_{1}, \dotsc, S_{k}$. Then, the maximal independent set contains vertices in $R$ corresponding to $S_{1}, \dotsc, S_{k}$ and pendant vertices corresponding to the remaining sets. It is easy to see that it is an independent set and it is also maximal. Moreover, the weight of the independent set is exactly $k$.
Similarly, you can prove the other direction, i.e., if there is a maximal independent set of value $k$, there is an independent set cover of size $k$.

Note: Instead of assigning $\infty$ value to the vertices in $L$, you can also assign some sufficiently large finite value, say $|\mathcal{R}|+1$. That will also work.