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. 

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.