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.]
Best Answer
It is easy to understand the existence of a Thom class by considering cellular cohomology. Let the given vector bundle be $E\to B$ with fibers of dimension $n$. One can assume without significant loss of generality that $B$ is a CW complex with a single 0-cell. The Thom space $T(E)$ is the quotient $D(E)/S(E)$ of the unit disk bundle of $E$ by the unit sphere bundle. One can give $T(E)$ a CW structure with $S(E)/S(E)$ as the only 0-cell and with an $(n+k)$-cell for each $k$-cell of $B$. These cells in $T(E)$ arise from pulling back the bundle $D(E)\to B$ via characteristic maps $D^k\to B$ for the $k$-cells of $B$. These pullback are products since $D^k$ is contractible.
In particular, $T(E)$ has a single $n$-cell and an $(n+1)$-cell for each 1-cell of $B$. There are no cells in $T(E)$ between dimension $0$ and $n$. The cellular boundary of an $(n+1)$-cell is 0 if $E$ is orientable over the corresponding 1-cell of $B$, and it is twice the $n$-cell in the opposite case. Thus $H^n(T(E);{\mathbb Z})$ is $\mathbb Z$ if $E$ is orientable and $0$ if $E$ is non-orientable. In the orientable case a generator of $H^n(T(E);{\mathbb Z})$ restricts to a generator of $H^n(S^n;{\mathbb Z})$ in the "fiber" $S^n$ of $T(E)$ over the 0-cell of $B$, hence the same is true for all the "fibers" $S^n$ and so one has a Thom class.