This left-right business is largely an accident of notation/language/writing, which makes it all the more difficult to distinguish genuine mathematical features from accidents of writing left-to-right horizonatally, etc.
Presumably a primary way in which $G^{op}$ arises is acting on the dual $V^\star$ of a given repn $V$ of $G$, by choosing to have $G$ act "on the right" on the dual, by $(\lambda g)(v)=\lambda(gv)$, that is, just moving the parentheses. The "problem", indeed, is the apparent left-right switch, if we write something like $\pi^\star(g)(\lambda)=\lambda\circ g$, namely, $\pi^\star(gh)=\pi^\star(h)\pi^\star(g)$.
It is a significant problem that $G^{op}$ [edit] cannot be isomorphic to $G$ non-abelian by $g\rightarrow g$... [as commented, my previous was messed-up...] Yes, it can be by $g\rightarrow g^{-1}$, but/and this amounts to converting everything back to left $G$-modules. My point would be that it is going the long way round to introduce the opposite group if we want to recover "left" $G$-modules in any case, and when the left-right business is syntactic, rather than semantic.
It is possibly better to have $G$ act "on the left" on the dual by $(g\lambda)(v)=\lambda(g^{-1}v)$.
Similarly, modules over rings can be "left" or "right", but, in fact, "right" modules are left modules over the opposite ring. Thus, a left $S$-module and right $R$-module is, as well, a left $S\otimes R^{op}$-module, where the left actions of $S$ and $R^{op}$ commute.
Some special classes of rings do have natural isomorphisms to their opposite rings, allowing identifications...
EDIT: yes, incorporating @Snark's comment, one common manner in which the left-right seeming-confusion arises is with a group $G$ acting on a set $X$, and then "inducing" a natural action of $G$ on functions $f$ on $X$. Much as taking dual reverses arrows, the action on functions on a set reverses arrows... thus the following. For $G$ acting on $X$ on the left, $G$ acts on functions "on the left" by $(gf)(x)=f(g^{-1}x)$. For $G$ acting on the right on $X$, $G$ acts on functions "on the left" by $(gf)(x)=f(xg)$. (Note that this avoids having to pretend that $G^{op}$ is a group, etc.) Once or twice in one's life, one should verify that the placement of inverse (or lack theoreof) makes the "left" action of $G$ on functions "associative", in the sense $(gh)f=g(hf)$ for $g,h\in G$.
Edit: ... and, in reference to the added "original question", operationally there is no "left/right" sense to a repn. One may choose different ways of writing, but the underlying things do not change. Many years ago, I. Herstein advocated writing functions on the right of their arguments, for the reason that then the order we'd read them would be the order of their application, but it mostly did not "catch on".
More edits! (Sorry for the delay... busy...) Certainly if one finds left-right distinctions/notations useful at least as a mnemonic, then they're good. For myself, somehow the variety of contexts relevant to my stuff makes me feel that left-right is insufficient. For one thing, different contexts would seem to impose conflicting requirements. For groups, many people do not want the contragredient to be a repn of the "opposite group", because "contragredient" is supposed to be a functor from the category of $G$-repns to itself. Similarly for special-but-important rings ("co-algebras", generally, I believe), such as group rings, universal enveloping algebras. "Oppositely", for rings without (relevant) involution, the bi-module, etc., notion is sometimes very useful. Nevertheless, it can get out of hand: for $R,S$-bimodule $M$ and $T,U$-bimodule $N$, $Hom_{\mathbb Z}(M,N)$ has a fairly obvious structure... until/unless we feel compelled to establish a once-and-for-all, "perfect" notation that allegedly suits all contexts.
I know that left-right is often used to suggest co/contra-variance, but, in fact, I claim that there are some technical advantages in not identifying the two things, one being innate, the other arguably less so.
Again, I'll claim that left-right should never be the pivotal issue, if the context is decently portrayed. Sure, something can be _tied_to_ notational left-right issues, but there's surely something underlying that is not so fragile as notation. This is not meant to "dis" anyone who finds left-right a marvelous device to remember distinctions!
Yet-another-edit: as with many things, the path by which one comes to a situation strongly affects one's attitude. Here, if one comes to "group repns" as a special and specially-equipped case of categories of modules over rings, yes, arguably, it makes complete sense to talk about the way in which categories over special types of rings have bonus properties. On the other hand, if one has followed the "opposite" route, starting with group repns as useful in various applications, and considered "imbedding" that story in the larger story of categories of modules over rings with-or-without various structures, one could just-as-easily take the viewpoint that the "general case" is a kind of "failure". That is, if categories of group repns are the "usual case" or "normal case", one might find no need to distinguish it from the "failing cases", and, likewise, not choose language that emphasizes how lucky we are to have "all these special features"... because they're not special?
Certainly some people prefer to proceed from the particular to the general, and others vice-versa, and such predilections color language and notation, obviously, and, often in manners nearly incomprehensible to people with "opposite" viewpoints.
Best Answer
I'm not sure whether a representation of an algebra $A$ means a representation of the unit group of $A$, or an $A$-module. With the second interpretation, the statement is false.
Let's take $k=2$ and use the negative definite inner product. (This example will occur inside any larger example.) So the Clifford algebra is generated by $e_1$ and $e_2$, subject to $e_1^2=e_2^2=1$ and $e_1 e_2 = - e_2 e_1$. Let $S$ be the $2$-dimensional representation $$\rho_S(e_1) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \rho_S(e_2) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ Let $V=S^{\otimes 3}$. I claim that $V$ is a $\mathrm{Pin}$ representation where $-1$ acts by $- \mathrm{Id}$, but $V$ is not a module for the Clifford algebra.
For all $\theta$, the vector $v(\theta) := \cos \theta e_1 + \sin \theta e_2$ is in the Pin group. Clearly, $$\rho_S( v(\theta)) = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}.$$ Then $\rho_V(v(\theta))$ is an $8 \times 8$ matrix I don't care to write down, whose entries are degree $3$ polynomials in $\sin \theta$ and $\cos \theta$. The point is, $$ \rho_V( v(\theta) ) \neq \cos \theta \rho_V(e_1) + \sin \theta \rho_V(e_2).$$ So $V$ is not an $A$-module. It is easy to build similar examples for the other signatures.
I'm not sure what happens if we read "representation of $A$" as "representation of the unit group of $A$."