The correct statement is: any 3-dimensional selfdual Galois representation is isomorphic to a quadratic twist of the adjoint of some 2-dimensional representation. (The quadratic twist is really necessary, as one can see from a variant of David Speyer's example.)
If $\rho: G_F \to GL_3(\overline{\mathbf{Q}}_\ell)$ is a Galois representation, where $F$ is a number field, and $\rho \cong \rho^*$, then the image of $\rho$ has to be contained in the subgroup $O_3(\overline{\mathbf{Q}}_\ell)$ of matrices preserving an orthogonal form. The quotient $O_3 / SO_3$ has order 2, so there is some quadratic character $\nu$ such that $\rho \otimes \nu$ takes values in $SO_3$.
There is an exceptional isomorphism $Ad: PGL_2 \cong SO_3$, so $\rho$ must be $Ad(\tau)$ for some 2-dimensional projective representation $\bar\tau$. You want to know if $\bar\tau$ lifts to an actual representation; the obstruction to this lies in some $H^2$, and by a theorem of Tate this vanishes, so $\bar\tau$ is the image in $PGL_2$ of some $\tau: G_F \to GL_2(\overline{\mathbf{Q}}_\ell)$.
(Note that $\tau$ is non-unique, and it is far from obvious a priori that $\tau$ can be chosen to be geometric if $\rho$ is. This kind of lifting-with-local-conditions question has been studied in detail in Patrikis' thesis.)
This statement has an analogue on the automorphic side: Theorem A of Ramakrishnan's paper "An exercise concerning the selfdual cusp forms on GL(3)" states:
Let F be a number field, and $\Pi$ a cuspidal, selfdual automorphic
representation of $GL_3(\mathbb{A}_F)$. Then there exists a non-dihedral cuspidal automorphic representation $\pi$ of $GL_2/F$, and an idele class character $\nu$ of F with $\nu^2 = 1$, such that $\Pi \cong Ad(\pi) \otimes \nu$.
EDIT. I just realised I had missed something: you are taking a (slightly unusual) definition of "selfdual". You aren't requiring that $\rho^* = \rho$ but just that $\rho^* = \rho \otimes \kappa^n$ for some $n$ (where $\kappa$ is the cyclotomic character), so $\rho$ is "selfdual up to Tate twists".
However if you make the even more general assumption that $\rho^* = \rho \otimes \chi$ for any character $\chi$ (this is what people call "essentially selfdual"), then comparing determinants we have $\chi^3 \det(\rho)^2 = 1$, and hence $\chi$ is a square in the group of characters of $G_F$. So we can twist $\rho$ to make it self-dual on the nose, and apply the previous argument to that; and the conclusion is still that $\rho$ is a twist of the adjoint of something (or, if you prefer, a twist of the symmetric square of something).
EDIT 2. (Let's not keep doing this, if you want to change the question then open a new question.) You now ask if any essentially selfdual 3-d representation is a quadratic twist of a symmetric square.
This is not clear to me. What is clear from the above arguments is the following: if $A_\ell$ is the abelian group of characters $G_F \to \overline{\mathbf{Q}}_\ell^\times$, and $\Sigma$ is any set of representatives for the quotient $A_\ell / 2 \cdot A_\ell$, then any ess. selfdual $\rho$ can be written as $\operatorname{Sym}^2(\tau) \otimes \chi$ for some 2-dimensional $\tau$ and some (unique) $\chi \in \Sigma$.
So your new question reduces to a purely one-dimensional one: is every class in $A_\ell / 2A_\ell$ represented by a quadratic character? But I don't see any reason why this should be true in general: I don't see why $A_\ell[2]$ should surject onto $A_\ell / 2A_\ell$.
