First, for any $x\in S^2$ we have an endomorphism $A(x)$ of $\mathbb{R}^3$ given by $A(x)(w)=x\times w$. More generally, we have an orthogonal matrix $B(t,x)=\exp(t A(x))$, which is a rotation through angle $t$ around $x$. When $w$ is perpendicular to $x$ we have $B(\pi/2,x)(w)=A(x)(w)$.
Let $F$ be the space of maps $f$ as in the question, and let $G$ be the space of maps $g:S^2\to S^2$ satisfying $g(-x)=-g(x)$ for all $x$. Given a function $f(x)=X(x)+u(x)x$, put $\phi(f)(x)=A(x)(X(x))+u(x)x=x\times X(x)+u(x)x$. This gives a homeomorphism $\phi:F\to G$. Using the maps $x\mapsto B(t,x)(X(x))+u(x)x$ (for $0 \leq t\leq \pi/2$) we see that $\phi(f)$ is homotopic to $f$ and so has the same degree. It is fairly standard that maps in $G$ have odd degree, and it follows that maps in $F$ have odd degree.
UPDATE: Romain asks about a generalisation for $n>2$. This seems harder. The analogous space $F$ is the space of sections of the bundle $W=n+2-L$ over $\mathbb{R}P^n$, where $L$ is the tautological bundle. By thinking about exterior algebras, one can produce an isomorphism $2^nL\simeq 2^n$. In the case $n=2$ this gives $4L\simeq 4$ so $W=4-L=4L-L=3L$, and by working out explicit formulae for this identification one is led to the proof above. If $n>2$ then $2^n>n+2$ so this approach will not work without some additional ideas.
