The answer is no. E.g., let 
$\alpha = -1,\beta = 0,f_0= 0,f_1= 0,f_2= -2,f_3= 1$. Then $f_{n+1} \leq \alpha f_n + \beta f_{n-1}$ for $n=1,2$, whereas 
$$
f_3 = 1 \not\leq 0= \left(\alpha ^2+\beta \right)f_1 +\alpha  \beta  f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}.
$$