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 $(1)$ for a triple $(\alpha, \beta, \gamma) \in \mathbb{F}_q^3$ holds if at least two of $\alpha, \beta$ and $\gamma$ are zero. A pair $(A, B)$ such that $\text{Tr}(A, B) = (\alpha, \beta, \gamma)$ generates a dihedral group in this case.
Condition $(2)$ holds if $\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta \gamma - 2 = 2$, the left-hand side being the *Fricke polynomial*. This corresponds to the case of an affine subgroup of $SL_2(\mathbb{F}_q)$.
Condition $(4)$ holds if $\alpha, \beta$ and $\gamma$ lie in a proper subfield of $\mathbb{F}_q$. The remaining conditions are presented in the Addendum below.
It follows from the claim that OP's decision problem boils down to:
> - deciding whether two elements of $\mathbb{F}_q$ are equal.
> - deciding whether $n$ elements of $\mathbb{F}_q$ ($n \le 4$) generate a proper subfield and *computing* the generated subfield.
> - deciding whether an element of $\mathbb{F}_q$ belongs to a given proper subfield $\mathbb{F}_{q'} \subset \mathbb{F}_q$.
The above claim follows from a theorem of Macbeath [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].
**Side note.**
Let $G$ be a finite group generated by $k$ elements.
Let $\varphi_k(G) = \frac{\vert \text{Epi}(F_k, G) \vert}{\vert \text{Hom}(F_k, G) \vert}$, where $F_k$ denotes the free group on $k$ generators. The ratio $\varphi_k(G)$ is the probability that two uniformly chosen random elements generate $G$.
By a theorem of Liebeck and Shalev, we have $\lim_n \varphi_2(G_n) = 1$ for any sequence of non-isomorphic finite simple groups, see [3, Theorem 1.1.1]. For a simple group $G_n(q)$ of Lie type of untwisted rank $n$, we have
$\varphi_2(G_n(q)) = 1 - O(\frac{n^3 \log^2(q)}{q^n})$ by results of Kantor, Lubotzky, Liebeck and Shalev, see [3, Theorem 1.1.2].
**Addendum.**
Let $Q(\alpha, \beta, \gamma) \Doteq \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta \gamma - 2$ be the *Fricke polynomial*.
Here are the five conditions derived by Daryl McCullough and Marcus Wanderley from Dickson's classification Theorem and Macbeath's Fricke polynomial criterion:
> $(1)$ *Dihedral case.* At least two of $\alpha, \beta$ and $\gamma$ are zero.
> $(2)$ *Affine case.* $Q(\alpha, \beta, \gamma) = 2$.
> $(3.A_4)$ The elements $\alpha, \beta$ and $\gamma$ lie in $\{-1, 0, 1\}$ and $Q(\alpha, \beta, \gamma) = 0$.
> $(3.S_4)$ The elements $\alpha, \beta$ and $\gamma$ lie in $\{\pm 1, \pm \sqrt{2}\}$ and $Q(\alpha, \beta, \gamma) = 1$, where $\sqrt{2}$ denotes one root of $X^2 - 2$.
> $(3.A_5)$ It's complicated! The authors redirect us to [2, Main Theorem (3)].
> $(4)$ *Affine or $SL_2(q')$, part 1.* The elements $\alpha, \beta$ and $\gamma$ lie in a proper subfield of $\mathbb{F}_q$.
> $(5)$ *Affine or $SL_2(q')$, part 2.* The integer $q$ is odd, $\alpha^2,\beta^2,\gamma^2$ and $\alpha \beta \gamma$ lie in a proper subfield $\mathbb{F}_{q'}$ of $\mathbb{F}_q$ and at least one of $\alpha, \beta$ and $\gamma$ does not lie in $\mathbb{F}_{q'}$.
The cited article (a very nice paper!) contains more details, and of course, the corresponding proofs.
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[1] D. McCullough, M. Wanderley, "Nielsen equivalence of generating pairs of $\text{SL}(2, q)$", 2013.
[2] D. McCullough, "Exceptional subgroups of $\text{SL}(2, F )$". Preprint available [here][4], accessed 4 January 2013.
[3] I. Pak, "What do we know about the product replacement algorithm?", 2000.
[4]: http://www.math.ou.edu/~dmccullough/research/manuscripts.html