Aloff--Wallach spaces were discussed in this question. They are quotients of SU(3) by U(1) indexed by a lattice of rank 2. Am I correct in guessing that the $K$-theory group $K_0$ of these spaces is the same for all elements in the lattice? (I am also assuming here, perhaps falsely, that the algebraic, topological, and smooth $K_0$ groups coincide.)
What is the case in the more general Eschenburg space setting?
Since it was not quite specified what sort of K-theory is relevant, I'll try to argue that topological complex K-theory is not the same for all Aloff–Wallach spaces.
The Eschenburg spaces are parametrized by $k=(k_1,k_2,k_3)$ and $l=(l_1,l_2,l_3)$ such that $k_1+k_2+k_3=l_1+l_2+l_3$ plus a gcd condition. The case $l_i=0$ should be the Aloff–Wallach spaces, and in this case the gcd condition means only that the $k_i$ are pairwise coprime.
The integral cohomology of Eschenburg spaces has been computed by Eschenburg, cf.
J.H. Eschenburg. New examples of manifolds with strictly positive curvature. Invent. Math. 66 (1982), 469–480.
J.-H. Eschenburg. Cohomology of biquotients. Manuscripta Math. 75 (1992), no. 2, 151–166.
The formulas for the additive structure can also be found in the thesis of Pongdate Montagantirud; it is given as follows, where $r=\sigma_2(k_1,k_2,k_3)=k_1k_2+k_1k_3+k_2k_3$ is an odd integer: $$ {\rm H}^i(X,\mathbb{Z})=\left\{\begin{array}{ll} \mathbb{Z} & i=0,2,5,7\\ \mathbb{Z}/r\mathbb{Z} & i=4\\ 0 & \textrm{otherwise}\end{array}\right. $$
Now we can compute topological K-theory using the Atiyah–Hirzebruch spectral sequence $$ E^{p,q}_2=H^p(X,KU^q(\ast))\Rightarrow KU^{p+q}(X). $$ with the differentials going $d_r:E^{p,q}_r\to E^{p+r,q-r+1}_r$. Note that KU is concentrated in even degrees and therefore the differentials can only be non-trivial for $q$ even and $r$ odd. Moreover, since the Atiyah–Hirzebruch sequence collapses rationally, the only nontrivial differentials can be the ones with $H^4(X,\mathbb{Z})\cong\mathbb{Z}/r\mathbb{Z}$ as target. Therefore, all differentials are trivial and the spectral sequence collapses at $E_2$. There is no extension problem involving the odd cohomology, so we find that $$ KU^n(X)=H^{\rm odd}(X,\mathbb{Z})\cong\mathbb{Z}^2 $$ for $n$ odd. There is one extension problem for even cohomology: however, the filtration would have smallest step $H^4$ and the two subsequent subquotients $\mathbb{Z}$. This is the direction in which extensions must necessarily split, so we find $$ KU^n(X)=H^{\rm ev}(X,\mathbb{Z})\cong \mathbb{Z}^2\oplus\mathbb{Z}/r\mathbb{Z} $$ for $n$ even.
Since there are Aloff–Wallach spaces with different $r$, there are Aloff–Wallach spaces whose topological K-theories differ (but only on the torsion subgroup in $KU^{2i}$). I don't know what would happen for $KO$. Probably the fact that the only torsion in the cohomology is odd could be exploited, but I think the calculations for $KO$ would be a lot messier.
Two parenthetical remarks:
I don't know much about smooth K-theory and what it would look like for Aloff–Wallach spaces, so I can't say anything about that.
I don't think that algebraic K-theory applies here. (I suppose that there are no natural lifts of Aloff–Wallach spaces to algebraic varieties; and I suppose that the Waldhausen K-theory of finite spaces over Aloff–Wallach manifolds has not been computed. The Waldhausen-type algebraic K-theory should split off a copy of the stable homotopy type of the underlying space, so these could also be distinguished by the torsion part $r$.)
