Does the following diagram commute? $$ \require{AMScd} \begin{CD} BU @>{\psi^k}>> BU \\ @VVV @VVV \\ BO @>{\psi^k}>> BO \end{CD} $$
Evidence for: $rc = 2$, it works for $BU(1) \rightarrow BSO(2)$ by looking at Chern classes, and I did a few manual computations in higher dimension.
Evidence against: it's not a ring map, let alone a $\lambda$-ring map.
I think the answer is yes, it commutes.
The identities $rc=2$, $\psi^k c=c\psi^k$, and additivity of these operations already implies that $\epsilon:= \psi^k r-r\psi^k$ satisfies $2\epsilon =0$ in $[BU,BO]$. So the answer is yes if you can show $[BU,BO]$ has no torsion; or equivalently that $KO^0(BU)=[BU,Z\times BO]$ has no torsion. Which I am sure is true, but I don't see a clean proof at the moment, or have a reference.
Note that there is the fiber sequence of spectra $\Sigma KO\xrightarrow{\eta} KO\xrightarrow{c} KU$, giving the "Bott exact sequence": $$ \cdots \to KU^{-1}(BU) \to KO^1(BU)\to KO^0(BU) \to KU^0(BU) \to \cdots. $$ We know that $KU^*(BU)$ is torsion free and concentrated in even degrees ($KU$ is complex orientable), so the claim that $KO^0(BU)$ is torsion free is equivalent to showing $$ KO^1(BU)=[BU, U/O] = 0. $$
The paper
Hara, Shin-ichiro(J-KYOT) Note on KO-theory of BO(n) and BU(n). J. Math. Kyoto Univ. 31 (1991), no. 2, 487–493. 55N15 (19L99)
shows that $KO^1(BU(n))=KO^{-7}(BU(n))=0$ by showing that $E_3$ of the Atiyah-Hirzebruch spectral sequence vanishes in the appropriate dimensions, and I'm sure the same argument should work with $BU(n)$ replaced by $BU$.
Here's a path to the same answer that riffs on your rank 1 result by using the splitting principle.
Fix a space $X$ and a rank $n$ complex vector bundle $E\rightarrow X$. The splitting principle buys us a space $f:Q\rightarrow X$ over which $f^{*}E$ is a direct sum of complex line bundles $\oplus L_{i}$ and $f^{*}$ is injective on $K^{*}$.
We have the maps $r$, $c$, and $\psi^{k}$, they all commute with $f^{*}$, and we've agreed on the result for line bundles. So then $$f^{*}\psi^{k}r[E]=\psi^{k}rf^{*}[E]=r\psi^{k}f^{*}[E]=f^{*}r\psi^{k}[E]$$
At first glance it appears we are faced with an equality in $KO(Q)$, and $f^{*}$ is not injective on $KO^{*}$, only $K^{*}$. But actually we have something better. Namely, your result shows that for any complex line bundle $L$, $r\psi^{k}L$ and $\psi^{k}rL$ agree $\textit{as oriented rank 2 real vector bundles}$. So the equality expressed above is actually represented by an isomorphism of real vector bundles $$\oplus_{i}\psi^{k}rL_{i}\rightarrow f^{*}r\psi^{k}E=\oplus_{i}r(L_{i}^{\otimes k})$$ The isomorphism carries one direct sum decomposition to the other. Since $SO(2)=U(1)$, the isomorphism is actually an isomorphism of complex vector bundles.
So the equality $f^{*}\psi^{k}r[E]=f^{*}r\psi^{k}[E]$ actually upgrades to one in Vect$_{\mathbb{C}}(Q)$ and therefore in $K(Q)$. So as long as $\psi^{k}r[E]$ is in the image of $r$, we are in good shape by injectivity of $f^{*}$ on $K^{*}$. And, although $r$ is not a ring map, as you mentioned, its image is closed under the ring operations, so we are in good shape after all.
