With the usual topology on $\Bbb R$, a compactification $\mathrm{id}_{\Bbb R}:\Bbb R\to v\Bbb R$ can have a remainder $v\Bbb R \setminus \Bbb R$ of cardinality $1,2, 2^{\aleph_0}=\mathfrak c,$ or $2^{\mathfrak c}.$ The only possibilities less than $\mathfrak c$ are $1,2.$
Suppose $\mathfrak c^+<2^{\mathfrak c}.$ What possible cardinals between $\mathfrak c$ and $2^{\mathfrak c}$ can be the cardinals of such remainders?
Is there perhaps a Forcing argument that can answer or partly answer this?
Every connected compact Hausdorff space of weight $\aleph_1$ is the remainder $v \mathbb R \setminus \mathbb R$ of some compactification of $\mathbb R$. In particular, $[0,1]^{\aleph_1}$ is the remainder of a compactification of $\mathbb R$, and therefore $\mathbb R$ has a compactification with remainder of cardinality $2^{\aleph_1}$.
Using forcing, one can show that it is consistent to have $\mathfrak{c} < 2^{\aleph_1} < 2^{\mathfrak{c}}$. (For example, Easton's Theorem immediately implies that we may get a model where $2^{\aleph_0} = \aleph_2$, $2^{\aleph_1} = \aleph_3$, and $2^{\aleph_2} = \aleph_4$, although Easton's Theorem is a bit overkill for this.) Thus it is consistent that $\mathbb R$ has a compactification with cardinality in $[\mathfrak{c}^+,2^{\mathfrak{c}})$.
The result about weight-$\aleph_1$ continua is proved by Dow and Hart in this paper. But the special case of $[0,1]^{\aleph_1}$ is actually much easier to prove, using the fact that $[0,1]^{\aleph_1}$ is separable. Let $\{d_1,d_2,d_3,\dots\}$ be a countable dense subset of $[0,1]^{\aleph_1}$. Map $\mathbb R$ into $[0,1] \times [0,1]^{\aleph_1}$ as follows. First map $\mathbb R$ onto the ray $[1,\infty)$, and then map $[1,\infty)$ into $[0,1] \times [0,1]^{\aleph_1}$ by linearly mapping each interval $[n,n+1]$ to the line segment connecting $(\frac{1}{n},d_n)$ to $(\frac{1}{n+1},d_{n+1})$ in $[0,1] \times [0,1]^{\aleph_1}$. This mapping embeds the ray $[1,\infty)$ in $[0,1] \times [0,1]^{\aleph_1}$, and its boundary in this embedding is precisely the set $\{0\} \times [0,1]^{\aleph_1} \approx [0,1]^{\aleph_1}$.
Edit: It is also possible to find a compactification $v \mathbb R$ of $\mathbb R$ such that $|v \mathbb R \setminus \mathbb R|$ has countable cofinality. In fact, I claim that the set $$T = \{|v\mathbb R \setminus \mathbb R| \,:\, v\mathbb R \text{ is a compactification of } \mathbb R \}$$ includes all cardinals of the form $2^\kappa$, where $\aleph_0 \leq \kappa \leq \mathfrak{c}$, and all countable limits of such cardinals. So, for example, in a model of set theory where $2^{\aleph_n} = \aleph_{\omega+n+1}$ for all $n$ (which is consistent, by Easton's Theorem), there is a compactification $v \mathbb R$ of $\mathbb R$ such that $|v \mathbb R \setminus \mathbb R| = \aleph_{\omega+\omega}$.
Lemma: Suppose $X$ is a connected compact Hausdorff space, and $X$ has a dense subspace $D$ that is both separable and path connected. Then there is a compactification of $\mathbb R$ whose remainder is (homeomorphic to) $X$.
Proof: The main ideas are already present in the third paragraph above. Let $\{d_1,d_2,d_3,\dots\}$ be a countable dense subset of $D$. Map $\mathbb R$ into $[0,1] \times X$ as follows. First map $\mathbb R$ onto the ray $[1,\infty)$, and then map $[1,\infty)$ into $[0,1] \times X$ by linearly mapping each interval $[n,n+1]$ to some path connecting $(\frac{1}{n},d_n)$ to $(\frac{1}{n+1},d_{n+1})$ in $[0,1] \times D$. This mapping embeds the ray $[1,\infty)$ in $[0,1] \times X$, and its boundary in this embedding is precisely the set $\{0\} \times X \approx X$.
My claim above follows almost immediately from this lemma. Each of the spaces $[0,1]^\kappa$, where $\aleph_0 \leq \kappa \leq \mathfrak{c}$, is separable and path connected, and so $|[0,1]^\kappa| = 2^\kappa \in T$ by the lemma.
To get countable limits of such cardinals, fix some infinite cardinals $\kappa_1,\kappa_2,\kappa_3,\dots \leq \mathfrak{c}$. Let $Y$ be the space obtained by gluing the endpoints of an interval to some (any) point of $[0,1]^{\kappa_1}$ and some (any) point of $[0,1]^{\kappa_2}$, gluing the endpoints of another interval to $[0,1]^{\kappa_2}$ and $[0,1]^{\kappa_3}$, gluing the endpoints of another interval to $[0,1]^{\kappa_3}$ and $[0,1]^{\kappa_4}$, and so on. (In other words, $Y$ is obtained by stringing together the $[0,1]^{\kappa_n}$ like beads on a necklace.) Finally, let $X$ be the one point compactification of $Y$. Then $X$ satisfies the hypotheses of the lemma, and $|X| = \sup_n 2^{\kappa_n}$.
Interestingly, this method seems hopeless for getting a compactification of $\mathbb R$ with a remainder of size $\aleph_\omega$. I wonder if this is possible?
