This is not a complete answer. The case $\beta=0$ is obvious, and we can assume wlog that $\beta>0$.
I relax slightly your conditions by looking for concave functions (and not necessarily strictly concave) to get a clearer idea of the solutions.
Let $g(x)=1-x$, so that $f(x)=(1-x)(1+x)^{\beta}$ with $f''(x)=\beta(1+x)^{\beta-2}(\beta-3-(\beta+1)x)$.
That gives a strictly concave $f$ for all $\beta<3$, and you can add to $g$ a very small concave and positive function to fulfill the strict condition for both $f$ and $g$.
For $\beta>3$, the function $(1-x)(1+x)^{\beta}$ has an inflection point at $\alpha=\frac{\beta-3}{\beta+1}$, with $f(\alpha)=\frac{4}{\beta+1}(\alpha+1)^{\beta}$ and $f'(\alpha)=2(\alpha+1)^{\beta-1}$, so you could replace the convex part on the left of $\alpha$ by a linear expression:
$$f(x)=2(\alpha+1)^{\beta-1}\left(x-\frac{\beta^2-6\beta+1}{(\beta+1)^2}\right)$$
provided that $f$ stays positive on $(0,1)$, i.e. $\beta<3+2\sqrt 2$.
Then modify $g$ on the same interval to satisfy the initial equation. That gives
$$g'(x)=2\frac{(\alpha+1)^{\beta-1}}{(1+x)^{\beta+1}}\left((1-\beta)x+\frac{\beta^3-5\beta^2+3\beta+1}{(\beta+1)^2}\right)$$
$$g''(x)=2\beta\frac{(\alpha+1)^{\beta-1}}{(1+x)^{\beta+2}}\left((\beta-1)x-\frac{2\beta^2-5\beta+3}{\beta+1}\right)$$
So the new expression of $g$ is concave for $x<\frac{2\beta^2-5\beta+3}{\beta^2-1}$. We still have to check:
$$\alpha<\frac{2\beta^2-5\beta+3}{\beta^2-1}\iff (\beta-3)(\beta-1)<2\beta^2-5\beta+3\\
\iff \beta^2-\beta>0$$
and this is true.
Finally, I can claim that you always have solutions for $\beta<3+2\sqrt 2$. I conjecture this is the upper bound, but I have no complete arguments.
