The answer is positive: there is a surjective, generically finite morphism $\text{sym}^d(C)\to \text{Prym}(C/C')$, at least away from small characteristics.  Fix a $k$-point $x$ of $C$, and use that to define an Abel map, $\alpha_x:C \to J(X)$.  The induced composition morphism $$C^g \xrightarrow{\alpha^g} J(C)^g \xrightarrow{\Sigma} J(C),$$ is surjective and generically étale.  Also the projection morphism, $$\pi:J(C) \to \text{Prym}(C/C')/\Gamma,$$ is a smooth morphism.  Since the Zariski tangent space of $C^g$ is generated by the Zariski tangent spaces of the $g$ fibers or the $g$ projections $C^g\to C^{g-1}$, up to a permutation of the factors, for sufficiently general $(x_{d+1},\dots,x_g)\in C^{g-d}$, the induced morphism $$C^d \times\{(x_{d+1},\dots,x_g)\} \to J(C) \to \text{Prym}(C/C')/\Gamma$$ is surjective and generically étale. Of course the morphism $$C^d \times \{(x_{d+1},\dots,x_g)\} \to J(C)$$ factors through the morphism $C^d \to \text{sym}^d(C)$ since the group law on $J(C)$ is Abelian.   Thus, in all, there is a surjective, generically étale morphism $$\psi:\text{sym}^d(C) \to \text{Prym}(C/C')/\Gamma.$$  Observe that, up to composing this morphism with a translation of $\text{Prym}(C/C')$, the morphism is independent of the choice of permutation or general point $(x_{d+1},\dots,x_g) \in C^{g-d}$.

Finally, the finite subgroup scheme $\Gamma$ is contained in the $N$-torsion subgroup scheme for some integer $N$ (that can be bounded just in terms of the topological data of $\phi$).  Thus, assuming the characteristic is larger than $N$, the multiplication by $N$ morphism, $$ \text{Prym}(C/C') \to \text{Prym}(C/C'),$$ is surjective and étale, and it factors through a surjective, étale morphism $$\chi:\text{Prym}(C/C')/\Gamma \to \text{Prym}(C/C').$$  Therefore, there is a surjective, étale morphism, $$\chi\circ \psi:\text{sym}^d(C) \to \text{Prym}(C/C').$$  

Of course that may not be the answer you want.  You still need to work out $\Gamma$ and $N$.  Also, "canonically" you only obtain a morphism from $\text{sym}^d(C)$ to a torsor for $\text{Prym}(C/C')$; trivializing the torsor depends on choosing a $k$-point of $C$ (at least as specified above).  So if you want to do this in families or over a non-algebraically closed field, you will need to do some work.  Finally, there is the question, in small characteristics, of what to do when the group scheme $\Gamma$ is not étale.