The rational cohomology of a homogeneous space $G/K$ admits a homomorphism from $H^*(BK)$ induced from the classifying map $G/K \to BK$ of the principal $K$-bundle $G \to G/K$. Assume the Lie group is $K$ connected, so that $\pi_1(BK) = 0$; then the space $G/K$ is formal in the sense of rational homotopy theory if and only if $H^*(G/K)$ is a free module over the image $A$ of the map $H^*(BK) \to H^*(G/K)$. In this event, one actually has a graded $A$-algebra decomposition $$H^*(G/K) \cong A \otimes \Lambda \hat P$$ for $\Lambda \hat P$ an exterior subalgebra on $\mathrm{rk \,}G - \mathrm{rk \,}K$ generators, isomorphic to the image of $H^*(G/K) \to H^*(G)$, and consequently the rational vector space dimension of $H^*(G/K)$ is given by $$\dim_{\mathbb Q} H^*(G/K) = (\dim_{\mathbb Q} A) \cdot 2^{\mathrm{rk \,}G - \mathrm{rk \,}K}.$$

The ring factorization above is too good to hold in general, but one can measure its failure with the deficiency, a constant defined as follows. In general the quotient map $G \to G/K$ still induces a map $H^*(G/K) \to H^*(G)$ whose image is an exterior algebra $\Lambda \hat P$, but the dimension of $\hat P$ is only bounded above by $\mathrm{rk \,}G - \mathrm{rk \,}K$, whereas in the formal case it is equal. The deficiency is the difference from this upper bound: $$\mathrm{df}(G/K) := \mathrm{rk \,}G- \mathrm{rk \,}K - \dim_{\mathbb Q} \hat P,$$ so that formality is deficiency $0$.

Now, it happens that if $\mathrm{df}(G/K) \in \{1,2\}$, then even though the ring structure on $H^*(G/K)$ is more recalcitrant than in the formal case, the dimension condition on $H^*(G/K)$ adduced above continues to hold.

Does this dimension condition hold in greater generality? If not, what's a counterexample?