If appropriately interpreted, the question does make sense. On the other hand, the answer by @TobiasDiez does not seem to be correct (a spinor cannot be expanded in general using a basis of *gamma matrices*). 

Let $S$ be a spinor bundle over a pseudo-Riemannian manifold $(M,g)$, whose bundle of Clifford algebras is denoted by $Cl(M,g)$. Since $S$ is a spinor bundle, there exists a morphism of bundles of associative, unital algebras:

$\gamma \colon Cl(M,g)\to End(S)$ 

to the endomorphisms of $S$. Let $\left\{ e_{\mu}\right\}$ be a local orthonormal frame of $TM$ over $U\subset M$. Using the injection $TM\subset Cl(M,g)$ we can apply $\gamma$ to  $\left\{ e_{\mu}\right\}$:

$\gamma_{\mu} := \gamma(e_{\mu})\in End_U(S)\, ,$

obtaining what the physicists usually call the *gamma matrices* $\gamma_{\mu}$. Therefore, the gamma matrices are nothing but local sections of the endomorphism bundle of the spinor bundle obtained by Clifford multiplication with respect to an orthonormal frame.

Now, let $\nabla^S$ be a connection on $S$. To the question "how does $\nabla^S$ acts on the gamma matrices?" the answer is simple: it acts through the connection induced by $\nabla^S$ on the endomorphism bundle of $S$ by the usual formula, since we just saw that the gamma matrices are local sections of the endomorphism bundle of $S$. A more interesting question would be: Is $\nabla^S$ a module derivation with respect to Clifford multiplication?, that is, is the following formula satisfied:

$\nabla^S (\gamma(x) (\xi)) = (\nabla^S \gamma(x)) (\xi) + \gamma(x) (\nabla^S  \xi) $

for all $x\in \Gamma(Cl(M,g))$ and $\xi\in \Gamma(S)$?. Here I am denoting by the same symbol the connection in $S$ and the induced connection on the endomorphism bundle. A more interesting situation happens when $\nabla^S$ is the *lift* of a metric connection $\nabla$ on $(M,g)$, which is the usual case in many applications. In this case, the natural question would be: is the following formula

$\nabla^S (\gamma(v) (\xi)) = ( \gamma(\nabla v)) (\xi) + \gamma(v) (\nabla^S  \xi) $

satisfied for all $v\in \mathfrak{X}(M)\subset \Gamma(Cl(M,g))$ and $\xi\in \Gamma(S)$? That is, we wonder if $\nabla^S$ is a module derivation with respect to Clifford multiplication and in addition:

$\nabla^S \gamma(v) = \gamma(\nabla v)$

for all $v\in \mathfrak{X}(M)$. If both these conditions are satisfied, $\nabla^S$ is usually called a *Clifford connection*. For the case of $S$ being a spinor bundle associated to a spin structure and $\nabla^S$ being the lift of the Levi-Civita connection $\nabla$, one can see that $\nabla^S$ is indeed a Clifford connection and thus:

$\nabla^S\gamma_{\mu} = \nabla^S\gamma(e_{\mu}) = \gamma(\nabla e_{\mu}) = \mathcal{A}_{\mu\nu} \otimes \gamma( e_{\nu}) = \mathcal{A}_{\mu\nu} \otimes \gamma_{\nu}$

where $\nabla e_{\mu} = \mathcal{A}_{\mu\nu} \otimes e_{\nu}$. Hence, the gamma matrices *behave* as vectors (or one-forms) with respect the Levi-Civita connection when applying $\nabla^S$ and this tells you how the "spin covariant derivative" $\nabla^S$ acts on gamma matrices in the case of a Clifford connection lifting the Levi-Civita connection, which is probably the situation of interest for the OP.