I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. I thought it more helpful to correct it than delete outright, but read critically.
If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.)
We can try to apply this to $Y=BG$, for $G$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out
in this recent answer,
all the torsion is actually 2-torsion.
It's sometimes worthwhile to consider the integral Stiefel-Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$.
[I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p_1$ is not a square.]
I want to say the short answer is no.
But in certain contexts, we can get things that are analogous. For example if you take a principal $B$-bundle $Q$ over $M$ and then suppose you can have a "nice" :) central extension of your lie group $B$ by $$1 \to \underline{\mathbb{C}^*}_M \to \tilde{B} \to B \to 1,$$where $\underline{\mathbb{C}^*}_M$ is the sheaf of smooth functions into $\mathbb{C}^*$, then you can define a cohomology class in $H^1(M, \underline{B})$ by seeing how well you can lift your bundle to a $\tilde{B}$-bundle. Now by the central extension, we would have $$H^1(M, \underline{B}) \overset{\sim}{=} H^2(M , \underline{\mathbb{C}^*}_M)$$ and then by the exponential sequence you would have $$H^1(M, \underline{B}) \overset{\sim}{=} H^2(M , \underline{\mathbb{C}^*}_M) \overset{\sim}{=} H^3(M, \mathbb{Z}).$$
And so what I am saying (since principal bundles have associated vector bundles) that your vector bundle in the right conditions could give you a degree three cohomology class in the integers. I'm sure you could pluck your Z/2 coefficients out of this (I don't know why you would want to be so rigid, nor am I claiming you are asking for that). Then there is an actual geometric interpretation of this integer related to a certain curvature form in this construction that I am not quite ready to add to this answer yet :)
Best Answer
It goes back to Wu in the 1950's that if one can compute the mod 2 cohomology of a manifold, with its Steenrod operations, then one can explicitly compute its Stiefel-Whitney classes, via the Wu formula. See for example the Theorem on page 188 of ``A concise course on algebraic topology'' (no originality claimed, just the quickest reference for me to find).