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The answer is no. E.g., let $\alpha = 2,\beta = 1,f_0= 0,f_1= 2,f_2= 4,f_3= 11$. Then $f_{n+1} \leq \alpha f_n + \beta f_{n-1}$ for $n=1,2$, whereas $$ f_3 = 11 \not\leq 10= \left(\alpha ^2+\beta \right)f_1 +\alpha \beta f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}. $$