The answer is yes, but only after tensoring with $\mathbb R$.

Thinking of the Beilinson regulator map with values in Deligne cohomology is simpler than thinking about the Borel regulator map; it's been proved that they agree with each other.

The topological Chern character map $ch_n : kU_{i} \to H^{2n-i}(pt,\mathbb Q)$ is an isomorphism $kU_{2n} \otimes \mathbb Q \ \xrightarrow \cong \ \mathbb Q$ when $i=2n$.

The corresponding algebraic Chern character map with values on Deligne cohomology is $ch_n : K_{i}\mathbb C \to H^{2n-i}_{\cal D}(pt,\mathbb Q(n))$. Here $\mathbb Q(n)$ (or $\mathbb Z(n)$) denotes a certain cochain complex of sheaves for the analytic topology on a complex manifold $X$. It starts in degree 0 with $\mathbb Q$ (resp., $\mathbb Z$), in cohomological degree 1 it has $\mathbb C$, and the differential map $d^0 : \mathbb Q \to \mathbb C$ is multiplication by $(2 \pi i)^ n$. The term in degree $i+1$ is the sheaf of holomorphic differentials $\Omega^i$ if $i < n$ and is $0$ if $i \ge n$. The exponential map $\mathbb C \to \mathbb C ^ \times $ given by $z \mapsto e^z$ gives a quasi-isomorphism $ \mathbb Z (1) \to \mathbb C ^ \times [-1]$; the degree shift there answers your second question, partially; another way of saying that is that there is a degree shift in the boundary map $c_1 : H^1(X,\mathbb C^\times) \to H^2(X,\mathbb Z)$. I say "partially", because one must know also that the regulator map involves no further degree shift; in degree 1 it's because the map $\mathbb C ^ \times \to \mathbb R$ given by $z \mapsto {\rm log} |z|$ involves no degree shift.

Now consider the projection $\mathbb C \to \mathbb R$ that sends $ (2 \pi i)^n $ to $0$ and $i^{n-1}$ to $1$; perhaps there is a better normalization for this map, such as choosing to send $(2 \pi i)^{n-1}$ to $1$. It induces a map of cochain complexes $\mathbb Q(n) \to \mathbb R[-1]$; the map it induces on Deligne cohomology, composed with the Chern character map above, is the Beilinson regulator map $$ch_n : K_{i}\mathbb C \to H^{2n-i}(pt,\mathbb R[-1]) = H^{2n-i-1}(pt,\mathbb R),$$whose only nonzero possibility is the map $ch_{n} : K_{2n-1}\mathbb C \to H^{0}(pt,\mathbb R) = \mathbb R$. For the ring of integers $A$, we get a map $K_{2n-1} A \to K_{2n-1}( A \otimes \mathbb C ) \to H^{0}(Spec(A \otimes \mathbb C),\mathbb R) = \mathbb R^{s+2t}$. Borel's theorem is recast as saying that for $n > 1$ the map induces an isomorphism of $(K_{2n-1} A) \otimes \mathbb R$ with the appropriate eigenspace for the action of $G = Gal(\mathbb C/\mathbb R)$ on $$H^0(Spec(A \otimes \mathbb C), \mathbb R (n-1)) = \mathbb R^{s+2t},$$where now I use $\mathbb R (n-1)$ to remind us how $G$ acts on this real vector space of dimension 1.

(**Added later**: actually, it may be more natural to replace the $\Sigma$ in the question by $\Omega$ and to use the anti-invariants (or invariants, depending on the parity of n) under $G$ acting on $K^{top}(A \otimes \mathbb C) \otimes \mathbb C$ instead of the invariants. Thus the degree shift can be viewed as $-1 = 1 - 2$)