Whitehead theorem for cohomotopy

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.

No, certainly not. If $X$ is a space with trivial reduced integral homology (for example the Poincare homology sphere with a ball removed) then the cohomotopy sets are all trivial, like those of a point.

EDIT Here are some details. Suppose $X$ has trivial homology, therefore trivial cohomology for all constant coefficients. Let $f:X\to Y$ be a based map with $Y$ connected and $\pi_1Y$ abelian. $Y$ is homotopy equivalent to a space that fibers over a $K(G,1)$ with $1$-connected fiber $Y_2$. $f$ can be lifted to a map $f_2:X\to Y_2$. $Y_2$ is homotopy equivalent to a space that fibers over a $K(G,2)$ with $2$-connected fiber $Y_3$. $f_2$ can be lifted to a map $X\to Y_{3}$, and this process can be repeated indefinitely. In the limit $f$ is lifted to an $\infty$-connected space, showing that it is homotopic to a constant.

Let me just add that stably it is true that a map f: $X\xrightarrow{} Y$ between finite complexes that induces isomorphisms on cohomotopy (of all degrees, now, including negative) is a weak equivalence. This is because the collection of all Z such that $f^{*} : [Y,Z] \xrightarrow{} [X,Z]$ is an isomorphism is a thick subcategory (also closed under products) so if it contains the spheres it contains all finite complexes. It is not true for general Y and Z because there are many many spectra with no cohomotopy at all; the Morava K-theories for example.

This also points to what goes wrong unstably, because mapping out of things behaves well with respect to fibrations, not cofibrations. So you'd expect the co-Whitehead theorem to be true not for finite complexes, but for spaces that can be built from spheres by taking iterated total spaces of fibrations; Lie groups for example, maybe.