Does there exist some constant $c > 0$ such that for all sufficiently large $n$, if $G$ is a graph with $n$ vertices and at least $(1/8 - c)n^2$ edges then $G$ must contain either a $K_4$ or an independent set on at least $n/\log n$ vertices?

The answer is no, as shown by Fox, Loh, and Zhao [FLZ15].

The result of Fox, Loh, and Zhao [FLZ15] disproving the problem: if $f(n) \geq 0$ satisfies $f(n) = o(\sqrt{\log n/\log\log n})$, then for every $\epsilon > 0$ and all sufficiently large $n$ there is a $K_4$-free graph on $n$ vertices with independence number at most $ne^{-f(n)}$ and at least $(1/8 - \epsilon)n^2$ edges; that is, $\mathrm{rt}(n; 4, ne^{-f(n)}) \geq (1/8 - o(1))n^2$. Applied with $f(n) = \log\log n$, this disproves the headline problem, since $ne^{-f(n)} = n/\log n = o(n)$.

The complementary result of Sudakov [Su03]: if $f(n)/\sqrt{\log n} \to \infty$ then $\mathrm{rt}(n; 4, ne^{-f(n)}) = o(n^2)$; that is, for every $\epsilon > 0$ and all sufficiently large $n$, every $K_4$-free graph on $n$ vertices with independence number at most $ne^{-f(n)}$ has at most $\epsilon n^2$ edges.

A sanity check for erdos_615: the empty graph on $n \geq 3$ vertices contains an independent set on at least $n/\log n$ vertices (namely the whole vertex set), so it satisfies the conclusion of the implication in the problem statement.