I found a proof for this statement. But the reason is another one that one might think. Let $C^-$ be the full subcategory generated by the objects of $C$ minus the $X$. I claim: If there is a simplex with more than one $X$-component at all, then both the inclusions $NC^-\rightarrow NC$ and $NC^-\rightarrow N^rC$ are homotopy equivalences. I give a very short explanation. First observe that the existence of a simplex with more than one $X$-component implies the existence of an object $A$ and arrows $A\rightarrow X$ and $X\rightarrow A$. From this one can show that the comma category $C^-\downarrow X$ is filtered and therefore contractible. Quillen's A implies that the inlusion $NC^-\rightarrow NC$ is a homotopy equivalence. On the other hand $N^rC$ is the pushout of
$$NC^-\leftarrow N((C^-\downarrow X) * (X\downarrow C^-))\rightarrow \operatorname{Cone}N((C^-\downarrow X) * (X\downarrow C^-))$$
(That's in fact the reason why I defined $N^rC$). So also $NC^-\rightarrow N^rC$ is a homotopy equivalence.

Probably the following is the upshot of this question: Whenever you see a category with an object $X$ with no non-trivial endomorphisms but an object $A$ with $A\rightarrow X$ and $X\rightarrow A$, then you can kick the $X$ and the homotopy type of the category doesn't change. This follows easily, as pointed out above, from Quillen's Theorem A which we all love so much.

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