Existence of a compactification of $\mathbb{R}$ with $\aleph_0$ remainder

I guess you are interested in Hausdorff compactifications. (It is easy to construct a non-Hausdorff compactification with 3-point remainder.)

Set $W=\beta\mathbb R\backslash \mathbb R$; note that $W$ has two connected components. If $Z$ is an other compactification of $\mathbb R$ then $V=Z\backslash \mathbb R$ is an image of $W$. Therefore $V$ has at most two connected components. It follows that cardinality of $V$ may be 1, 2, or something between $\mathfrak{c}$ and $|W|=2^{\mathfrak{c}}$.

EDIT: As Emil pointed out, I misinterpreted a statement in Charalambous article. Corrections are in boldface:

I don´t know if The answer to A is yes. For a separable metric space X (e.g. $\mathbb{R}$) the following are equivalent (see Compactifications with countable remainder by Charalambous in PAMS 1980):

1. X is Cech-complete and rim-compact.

2. X has a compactification with at most countable remainder.

$\mathbb{R}$ is Cech-complete because it is complete, and it is rim-compact because it is locally compact.

So $\mathbb{R}$ has a compactification with at most countable remainder, but you already knew that.