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 $\phi(f)(x)=A(x)(X(x))+u(x)x=x\times A(x)+u(x)x$.  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.