No, this is false. According to the Sullivan Conjecture (Miller's Theorem), $\mathrm{map}_*(B\mathbb{Z}/p, S^n) \sim *$ for all $n$, which means $$ [\Sigma^n B\mathbb{Z}/p, S^k] = * $$ for all $n$. So if we let $f: \Sigma^k \mathbb{Z}/ p \to *$, the induced map $$ f^*: \pi^k(*) \to \pi^k ( \Sigma^n B\mathbb{Z}/p ) $$ is the equivalence $* \to *$. Since $f$ is not a homotopy equivalence, this counterexamps the conjecture.
Perhaps it would be more interesting to restrict attention to maps $f:X\to Y$ between finite complexes.
EDIT (further thoughts): If $K$ and $L$ are finite complexes, then something like your co-Whitehead statement is true!
Theorem 1: If $f: K\to L$ is a map of finite complexes such that $\pi^k( \Sigma^n f)$ is an isomorphism for all $k\geq k_0$ and all $n \geq n_0$, then $\Sigma f$ is a homotopy equivalence.
Corollary 2: In Theorem 1, if both $K$ and $L$ are simply-connected, then $f$ is a homotopy equivalence.
The proof uses a theorem of mine:
Theorem M: If $X$ is simply-connected and of finite type and $\mathrm{map}_*(X,S^k) \sim *$ for all sufficiently large $k$, then $\mathrm{map}(X,Y)\sim *$ for all finite-dimensional CW complexes $Y$.
Proof of Theorem 1: The hypotheses imply that the cofiber $C_{\Sigma^{n_0} f} \simeq \Sigma^{n_0} C_f$ satisfies $\mathrm{map}_*(\Sigma^{n_0} C_f, S^k) \sim *$ for all $k \geq k_0$. Theorem M implies that $\mathrm{map}_*(\Sigma^{n_0} C_f, \Sigma^{n_0}C_f) \sim *$, which implies $\Sigma^{n_0} C_f \sim *$ and hence that $\Sigma C_f \sim *$. This suffices to show that $\Sigma f$ is a homotopy equivalence.