The answer is **yes**, it is *easy to check* whether the ordered pair $(A, B)$ generates $SL_2(\mathbb{F}_q)$ for $q$ the power of a prime. Indeed, there exist *simple criteria* according to Daryl McCullough and Marcus Wanderley, see [1, Section 11]. (*Italic* means that I am quoting the authors).

>> **Claim.**
The pair $(A, B)$ generates $SL_2(\mathbb{F}_q)$ if and only if
 $$\text{Tr}(A, B) \Doteq (\text{Tr}(A), \text{Tr}(B), \text{Tr}(AB))$$ is an essential triple, i.e.,
doesn't satisfy any of the conditions $(1) - (5)$ of [1, Section 11].
 
For instance, the condition $(2)$ for a triple $(\alpha, \beta, \gamma) \in \mathbb{F}_q^3$ holds if $\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta \gamma - 2 = 2$ (the left-hand side is the Fricke polynomial). This corresponds to the case of an affine subgroup of $SL_2(\mathbb{F}_q)$. 

The above claim follows from a Theorem of Macbeath, namely [1, Theorem 8.2] while the criteria are based on Dickson's Theorem [1, Theorem 6.1] and another theorem of Macbeath [1, Theorem 8.1].

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[1] D. McCullough, M. Wanderley, "Nielsen equivalence of generating pairs of $\text{SL}(2, q)$", 2013.