Stiefel-Whitney classes are defined (for example, in Milnor's Characteristic classes) as elements of the cohomology groups with $\mathbb{Z}_2$-coefficients as follows. $$w_i(E)=\phi^{-1} Sq^i(u)$$ where $E$ is an $n$-plane bundle over $B$, $E_0$ is the complement of the zero section, $Sq^i: H^*(E,E_0)\to H^{*+i}(E,E_0)$, $u\in H^n(E,E_0)$ is the Thom class and $\phi: H^*(B)\to H^{*+n}(E,E_0)$ is the Thom isomorphism (using $\mathbb{Z}_2$ coefficients).
My question is: if $E$ is oriented, does a similar construction work with $\mathbb{Z}$-coefficients? If yes, does it have a name (except of the "top" Euler class)? The only problem I see is with the Steenrod square operation which may be undefined for integral coefficients. If there are problems with the Steenrod squares, could such classes be defined inductively similarly as Chern classes for complex bundles?
You might be interested in the definition of both Stiefel-Whitney and Chern Classes in terms of the outer tensor product with $\gamma^1 \to RP^\infty$ and $\gamma^1 \to CP^\infty$ resp. This is a refreshing different approach to show the existence of those classes. You can find it for example briefly in Ebert's script about cobordism.
Note that you use the nice cohomology rings of those base spaces for the definition. Also note that this definition uses line bundles, which are boring in the orientable case, so the canonical idea might not work.