*perhaps he is implying some even stronger result*

He is referring to the following result of Peter Freyd ([Freyd uncertainty principle][1]):

The homotopy category of spaces $HoTop$ does not admit a faithful functor to the category of sets $Set$. Specifically, for any functor $T: Top_* \to Set$ from base-pointed spaces to sets which is homotopy invariant, there exist a triple $f: X \to Y$ such that $f$ is not null-homotopic, but $T(f) = T(\ast)$. Here $\ast$ is the null map to the basepoint of $Y$.

In particular, any algebraic invariant is a set-valued homotopy invariant. This includes homotopy groups, cohomology, cohomology and homotopy operations and whatever you can think of. Freyd's theorem implies that we cannot describe the homotopy category as a category of algebras for some algebraic theory $\mathcal{T}$, since any algebraic category is concrete.

Fun fact: Freyd's theorem essentially relies only on the general set-theoretic arguments and cardinal counting.

Non-counter-example: Whitehead's theorem states that if a map $f: X \to Y$ between pointed connected CW-complexes induces an isomorphism on all homotopy groups $\pi_i, \ i=1, 2,\dots$, then $f$ is a homotopy equivalence between $X$ and $Y$. Note that you still can't discriminate spaces looking just at the collection of homotopy groups: there can be $X$ and $Y$ such that $\pi_i(X) = \pi_i(Y)$ for all $i$, but this isomorphism is not induced by any actual map $F: X\to Y$ and the spaces are not homotopy equivalent. The simplest example is $\Bbb R \Bbb P^2 \times S^3$ and $S^2 \times \Bbb R \Bbb P ^3$. They both have a double covering by $S^2 \times S^3$ and have thus the same homotopy groups, but their cohomology is not isomorphic.

On second thought, I can't see why Freyd's theorem would imply actual non-discriminable spaces without any extra conditions on the invariant. Perhaps someone can fill this gap, but imho the non-discrimination of maps is bad enough.

  [1]: http://www.tac.mta.ca/tac/reprints/articles/6/tr6abs.html