The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $G$ with cyclic Sylow subgroups as being Sylow-cyclic. One can show that such groups are in fact metacyclic in the above sense, but I wont use this.

The answer to the OP's question is yes, in fact the following holds: Let $G$ be Sylow-cyclic and $M$ a finitely generated $\mathbb{Z}G$-module. If $P$ denotes a Sylow $\ell$-subgroup for the largest prime divisor $\ell$ of $\lvert G\rvert$, then $i_{P,*}\colon H_k(P,M)\rightarrow H_k(G,M)$ is surjective for any $k\geq 1$.

This as a consequence of the following observation (used repeatedly): Let $G$ be any finite group and $M$ and $k$ as above. If $p$ is the smallest prime divisor of $\lvert G\rvert$ and the Sylow $p$-subgroup $S$ is cyclic, then there exist a normal subgroup $N$ of order $[G:S]$ (a normal $p$-complement). Furthermore $i_{N,*}\colon H_k(N,M)\rightarrow H_k(G,M)$ is surjective. The first part follows from Burnsides transfer theorem (e.g. see Huppert, Endliche Gruppen I, Satz IV.2.7). The second follows from the Lyndon-Hochild-Serre spectral sequence corresponding to the short exact sequence $1\rightarrow N\rightarrow G\rightarrow S\rightarrow 1$. Since $N$ and $S$ have coprime orders, the spectral sequence collapses on the $E^2$-page and hence we have the surjection $H_k(N,M)\twoheadrightarrow H_0(S,H_k(N,M)) \stackrel{\cong}{\rightarrow} H_k(G,M)$.

The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $G$ with cyclic Sylow subgroups as being Sylow-cyclic. One can show that such groups are in fact metacyclic in the above sense, but I wont use this.

The answer to the OP's question is yes: For $k\geq 1$, $H_k(G,M)$ is a finite abelian group so we can write it as a sum of it’s $p$-torsion parts $H_k(G,M) = \bigoplus_p H_k(G,M)_{(p)}$. If $S$ is a Sylow $p$-subgroup the composite $H_k(G,M)\stackrel{\text{tr}}{\rightarrow} H_k(S,M)\stackrel{i_{S,*}}{\rightarrow} H_k(G,M)$ is multiplication by $[G:S]$. Looking at the $p$-torsion part the composite becomes an isomorphism, so $i_{S,*}\colon H_k(S,M) \rightarrow H_k(G,M)_{(p)}$ is surjective.

Thus $H_k(G,M)$ is always generated by the images corresponding to the Sylow subgroups. If these are all cyclic then $H_k(G,M)$ is generated by images corresponding to cyclic subgroups.

The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $G$ with cyclic Sylow subgroups as being Sylow-cyclic. One can show that such groups are in fact metacyclic in the above sense, but I wont use this.

The answer to the OP's question is yes, in fact the following holds: Let $G$ be Sylow-cyclic and $M$ a finitely generated $\mathbb{Z}G$-module. If $P$ denotes a Sylow $\ell$-subgroup for the largest prime divisor $\ell$ of $\lvert G\rvert$, then $i_{P,*}\colon H_k(P,M)\rightarrow H_k(G,M)$ is surjective for any $k\geq 1$.

This as a consequence of the following observation (used repeatedly): If $p$ is the smallest prime divisor of $\lvert G\rvert$ and $S$ is a Sylow $p$-subgroup, then there exist a normal subgroup $N$ of order $[G:S]$ (a normal $p$-complement). Furthermore $i_{N,*}\colon H_k(N,M)\rightarrow H_k(G,M)$ is surjective. The first part follows from Burnsides transfer theorem (e.g. see Huppert, Endliche Gruppen I, Satz IV.2.7). The second follows from the Lyndon-Hochild-Serre spectral sequence corresponding to the short exact sequence $1\rightarrow N\rightarrow G\rightarrow S\rightarrow 1$. Since $N$ and $S$ have coprime orders, the spectral sequence collapses on the $E^2$-page and hence we have the surjection $H_k(N,M)\twoheadrightarrow H_0(S,H_k(N,M)) \stackrel{\cong}{\rightarrow} H_k(G,M)$.

The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $G$ with cyclic Sylow subgroups as being Sylow-cyclic. One can show that such groups are in fact metacyclic in the above sense, but I wont use this.

The answer to the OP's question is yes, in fact the following holds: Let $G$ be Sylow-cyclic and $M$ a finitely generated $\mathbb{Z}G$-module. If $P$ denotes a Sylow $\ell$-subgroup for the largest prime divisor $\ell$ of $\lvert G\rvert$, then $i_{P,*}\colon H_k(P,M)\rightarrow H_k(G,M)$ is surjective for any $k\geq 1$.

This as a consequence of the following observation (used repeatedly): Let $G$ be any finite group and $M$ and $k$ as above. If $p$ is the smallest prime divisor of $\lvert G\rvert$ and the Sylow $p$-subgroup $S$ is cyclic, then there exist a normal subgroup $N$ of order $[G:S]$ (a normal $p$-complement). Furthermore $i_{N,*}\colon H_k(N,M)\rightarrow H_k(G,M)$ is surjective. The first part follows from Burnsides transfer theorem (e.g. see Huppert, Endliche Gruppen I, Satz IV.2.7). The second follows from the Lyndon-Hochild-Serre spectral sequence corresponding to the short exact sequence $1\rightarrow N\rightarrow G\rightarrow S\rightarrow 1$. Since $N$ and $S$ have coprime orders, the spectral sequence collapses on the $E^2$-page and hence we have the surjection $H_k(N,M)\twoheadrightarrow H_0(S,H_k(N,M)) \stackrel{\cong}{\rightarrow} H_k(G,M)$.