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 $A$."