Is there the Whitehead theorem for cohomology theory? We know that there are Whitehead theorem for homotopy and homology theory.
Is there the Whitehead theorem for cohomology theory for 1-connected CW complexes?
 A: Sure.
The basic point is that for simply-connected spaces, you can determine the connectivity of a map by looking at the connectivity of the cofiber instead of the connectivity of the fiber. 
In homology, you determine the connectivity of the cofiber by looking at $H_*(C;\mathbb{Z})$, because of the Hurewicz Theorem.  
In cohomology, you appeal to: if $X$ is simply-connected, then $X$ is $n$-connected if and only if $[ X, K(G,m)] = *$ for all $m \leq n$ and all abelian groups $G$.  This is because (by basic obstruction theory) if $X$ is $(n-1)$-connected, then $[X, K(G,n)] \cong \mathrm{Hom}(\pi_n(X), G )$. 
This results in a long list of groups to check, admittedly; but it can be whittled down by Universal Coefficients theorems if you like.
A: Conceptually, the following two theorems (both due to Whitehead) are Eckmann-Hilton duals.
Theorem.  A weak homotopy equivalence between CW complexes is a homotopy equivalence.
Theorem.  A homology isomorphism between simple spaces is a weak homotopy equivalence.
They don't look dual, but they are.  See
J.P. May. 
The dual Whitehead Theorems.
London Math. Soc.  Lecture Note Series 
Vol. 86(1983), 46--54.
The point is that the second statement is really about cohomology, and the standard
cellular proof of the first statement dualizes word-for-word to a ``cocellular'' proof
of the second.  Cocellular constructions are what appear in Postnikov towers, and they
can be used more systematically than can be found in the literature.  Yet another plug:
they are central in the upcoming book "More Concise Algebraic Topology'' by Kate Ponto
and myself.  
A: As Sean says, the key point is the Universal Coefficient Theorem, but the details are not completely obvious unless you make some finiteness assumptions.
Suppose that $f:X\to Y$ is such that $H^{\ast}(f;\mathbb{Z})$ is an isomorphism.  Let $Z$ be the cofibre of $f$, so $\tilde{H}^{\ast}(Z;\mathbb{Z})=0$.  If we can prove that $\tilde{H}_{\ast}(Z;\mathbb{Z})=0$ then we can appeal to the ordinary homological Whitehead theorem.  MathJax is mangling my tildes: all (co)homology groups of $Z$ below should be read as reduced. 
For any prime $p$, we have a universal coefficient sequence for $H^{\ast}(Z;\mathbb{Z}/p)$ in terms of $H^{\ast}(Z;\mathbb{Z})$, so $H^{\ast}(Z;\mathbb{Z}/p)=0$.  As ${\mathbb{Z}/p}$ is a field we also know that $H_{\ast}(Z;\mathbb{Z}/p)$ is a free module with $H^{\ast}(Z;\mathbb{Z}/p)$ as its dual, so we must have $H_{\ast}(Z;\mathbb{Z}/p)=0$.  Using the universal coefficient theorem for homology we deduce that $H_{\ast}(Z;\mathbb{Z})/p$ and $\text{ann}(p,H_{\ast}(Z;\mathbb{Z}))$ are zero, so multiplication by $p$ is an isomorphism on $H_{\ast}(Z;\mathbb{Z})$.  As this holds for all $p$, we see that $H_{\ast}(Z;\mathbb{Z})$ is a rational vector space.  Thus, if it is nontrivial it will contain a copy of $\mathbb{Q}$ so (via universal coefficients again) $H^{\ast}(Z;\mathbb{Z})$ will contain a copy of $\text{Ext}(\mathbb{Q},\mathbb{Z})$.  This group is nonzero (in fact, enormous) by a standard calculation, so this contradicts the initial assumption.
