Does every non-type-I factor's projection lattice admit a dense embedding of the standard continuum-collapsing poset?

Let $$R$$ be a non-type-I factor acting on a separable Hilbert space.

Let $$P(R)$$ be the set of $$R$$'s projections with the usual ordering ($$x \leq y \iff$$ range$$(x) \subseteq$$ range$$(y)$$) under which it forms a complete lattice; let $$P^+(R)$$ be the same excluding the null projection.

Let $$C$$ be the poset of finite sequences of ordinals $$< 2^\omega$$, ordered by reverse inclusion (note $$C$$ is the most common poset used in forcing to collapse $$2^\omega$$ to $$\omega$$).

Question: Must there exist an order-embedding $$\phi : C \rightarrow P^+(R)$$ whose range is order-dense in the sense relevant to forcing (namely for all $$x \in P^+(R)$$ there exists $$c \in C$$ such that $$\phi(c) \leq x$$)?

I'm not sure how much intrinsically operator-algebraic interest this question holds, but it is potentially interesting for set theory, in particular for forcing. If there is such a embedding then $$P^+(R)$$ is forcing-equivalent to the standard continuum-collapse forcing; if not, I suspect $$P^+(R)$$ would (at least for some types of factor $$R$$) be inequivalent to known forcing posets, and would presumably have some novel properties.

Note that there definitely exist order-embeddings of $$C$$ into $$P^+(R)$$ that map each unbounded descending chain in $$C$$ to an unbounded descending chain in $$P^+(R)$$, but this property by itself doesn't ensure the embedding is dense.

I believe I've shown (see answer to related stackoverflow question at https://math.stackexchange.com/a/4223869/250373 ) that there is such an embedding, on the supposition that every nontrivial lattice partition of $$P(R)$$ has continuum cardinality. This supposition is probably true in general but for now I can only prove this in the type III case, and with the assumption that CH holds.