Your question is not well written and has several possible interpretations. Here is one:
Does there exist a subgroup $G$ of $R^3\rtimes O(3)$ (containing no translations), a point $p\in R^3$ and a number $r>0$ such that for every $q\in R^3$ the intersection $B(q,r)\cap G\cdot p$ is finite and nonempty?
Lemma. Such subgroups do not exist.
Proof. Suppose that $G$ is such a group. Then $G$ has to be discrete (the finiteness assumption on $B(q,r)\cap G\cdot p$). If $G$ is crystallographic, i.e., $R^3/G$ is compact, then $G$ contains nontrivial translations. The alternative is that $G$ either has an invariant line $L$ or fixed point in $R^3$. Consider the first case. Then $G\cdot p$ is contained in $R$-neighborhood of $L$. Therefore, for arbitrary $r>0$, take $q$ such that $d(q, L)>r+R$. Then $B(q,r)$ contains no points from $G\cdot p$.
If $G$ has a fixed point $a$, then $G\cdot p$ is contained in the $R$-ball $B(a,R)$ for some $R$. Now, take $q$ such that $d(q,a)>R+r$. Again a contradiction. qed
A side remark: Every group of isometries of $R^3$ which contains no translations either fixes a point in $R^3$ or has an invariant line in $R^3$. One can ask about groups which contain no skew translations (isometries without fixed points). Then such subgroup of isometric motions of $R^3$ has to fix a point in $R^3$. Hence, it is conjugate to a subgroup of $O(3)$. Interestingly enough, there are isometry groups of $R^4$ which contain no (skew) translations but are not conjugate to subgroups of $O(4)$. (A proof that I know uses the fact that $U(2)$ contains free nonabelian subgroups.)