Allow me to say something which is not so much an answer to this question as to a (very natural) question that I sense is coming in the future.
There are two possible pitfalls in the definition of "has complex multiplication" for abelian varieties over an arbitrary ground field $F$.
1) Let's first talk only about complex abelian varieties. There is a serious discrepancy between the terminology "has complex multiplication" as used in the classical literature (up to and including, say, some of Shimura's papers in the 1960s) and the way it is almost invariably used today.
Classically, an abelian variety over the complex numbers was said to "have complex multiplications" whenever its endomorphism ring was strictly larger than $\mathbb{Z}$. This is never the generic case, but starting in dimension $2$ it allows abelian surfaces with multiplication by an order in a real quadratic field (parameterized by Hilbert modular surfaces), by an order in an indefinite rational quaternion algebra (parameterized by Shimura curves), and so forth.
Nowadays one means something much more restrictive by "complex multiplication": one wants the endomorphism ring to be an order in a $\mathbb{Q}$-algebra whose dimension is large compared to the dimension, say $g$, of $A$. In the case of an abelian variety without nontrivial abelian subvarieties ("simple"), CM means precisely that the endomorphism ring is an order in a number field of degree $2g$: it then follows from this (not so obviously) that the number field is totally complex and has an index $2$ totally real subfield. For nonsimple abelian varieties, the generally agreed upon definition is that $\operatorname{End}(A)$ should contain an order in a commutative semisimple $\mathbb{Q}$-algebra of dimension $2g$. Such things do not vary in moduli, even in higher dimensions.
2) If you are interested in abelian varieties over an arbitrary ground field $F$ (let me concentrate on the case of characteristic 0), then you need to distinguish between the ring of endomorphisms which are rationally defined over $F$ and the ring of endomorphisms which are defined over an algebraic closure of $F$ (hence, it turns out, over any algebraically closed field containing $F$, i.e., one cannot acquire endomorphisms by passing from $\overline{\mathbb{Q}}$ to $\mathbb{C}$). If one does not specify, "having CM" means having CM over the algebraic closure.
As in the responses above, it is very often the case that you have to extend the ground field slightly in order to get all the endomorphisms rational over the ground field. E.g. if $E_{/\mathbb{C}}$ is any elliptic curve, it can be minimally defined over $\mathbb{Q}(j(E))$. If $E$ has complex multiplication (over $\mathbb{C}$), then $j(E)$ is a real [not necessarily totally real] algebraic number, from which it follows that $\operatorname{End}_{\mathbb{Q}(j(E))}(E) = \mathbb{Z}$.
In particular, the $\ell$-adic Galois representation of a "CM abelian variety" [using standard conventions as in 1) and 2) above] need not have abelian image, quite. It has "almost abelian image", meaning that once you make a finite extension of the ground field to make all the endomorphisms rational, then the image will be abelian. (Both M. Emerton's and my arguments in the one-dimensional case extend easily to this case, although I think his is more insightful.)
What I have said carries over to fields of positive characteristic, but the general picture of endomorphism algebras looks different, especially over finite fields: every abelian variety over a finite field has complex multiplication in the above sense.
(EDIT: I've rewritten my argument in terms of the inverse functor, i.e., base extension, since it is clearer and more natural this way.)
Much of what is below is simply a reorganization of what Robin Chapman wrote.
Theorem: For each prime $p$, the base extension functor from the category $\mathcal{C}_{-p}$ of elliptic curves over $\mathbf{F}_{p^2}$ on which the $p^2$-Frobenius endomorphism acts as $-p$ to the category of supersingular elliptic curves over $\overline{\mathbf{F}}_p$ is an equivalence of categories.
Proof: To show that the functor is an equivalence of categories, it suffices to show that the functor is full, faithful, and essentially surjective. It is faithful (trivially), and full (because homomorphisms between base extensions of elliptic curves in $\mathcal{C}_{-p}$ automatically respect the Frobenius on each side). Essential surjectivity follows from Lemma 3.2.1 in
Baker, González-Jiménez, González, Poonen, "Finiteness theorems for modular curves of genus at least 2", Amer. J. Math. 127 (2005), 1325–1387,
which is proved by constructing a model for one curve and getting models for the others via separable isogenies. $\square$
The same holds for the category $\mathcal{C}_p$ defined analogously, but with Frobenius acting as $+p$.
Here are two approaches for proving essential surjectivity for $\mathcal{C}_p$:
1) If $G:=\operatorname{Gal}(\overline{\mathbf{F}}_p/\mathbf{F}_{p^2})$ and $E$ is an elliptic curve over $\mathbf{F}_{p^2}$, and $\overline{E}$ is its base extension to $\overline{\mathbf{F}}_{p^2}$, then the image of the nontrivial element under $H^1(G,\{\pm 1\}) \to H^1(G,\operatorname{Aut} \overline{E})$ gives the quadratic twist of $E$ (even when $p$ is $2$ or $3$, and even when $j$ is $0$ or $1728$). Applying this to each $E$ with Frobenius $-p$ gives the corresponding elliptic curve with Frobenius $+p$.
2) Use Honda-Tate theory (actually, it goes back to Deuring in this case) to find one supersingular elliptic curve over $\mathbf{F}_{p^2}$ with Frobenius $+p$, and then repeat the proof of Lemma 3.2.1 to construct the models of all other supersingular elliptic curves via separable isogenies.
Best Answer
There is nothing wrong. A typical case will be $\mathbb F_q = \mathbb F_{p^2}$ and $\varphi = -p$. Since $q^{1/2} = (p^2)^{1/2} = p$, this is consistent with RH.