TITLE: Pointed vs. unpointed homotopy colimits
QUESTION [6 upvotes]: Let $C$ be a category with a zero object, i.e. an object 0 which is both initial and terminal.  Then $C$ is automatically (and uniquely) enriched over the category $Set_\star$ of pointed sets with smash product, where the basepoint $0\in C(x,y)$ is the unique map which factors through the object 0.  Also, a functor between such categories is $Set_\star$-enriched (i.e. preserves the basepoints of homsets) if and only if it preserves the zero object.  Call these pointed categories and pointed functors.  (This is the "zero-ary" version of additive categories and biproducts.)
Now let $G:C^{op} \to Set_\star$ and $F:C\to Set_\star$ be pointed functors; we can then consider the tensor product $G \otimes_C F$ (otherwise known as the $G$-weighted colimit of $F$) either (1) as ordinary unenriched functors, or (2) as $Set_\star$-enriched functors.  The first is a quotient of $\bigvee_c G(c)\times F(c)$, while the second is a quotient instead of $\bigvee_c G(c) \wedge F(c)$.  (I'm writing $\vee$ for the coproduct in $Set_\star$, since it is a "wedge" which identifies basepoints in the disjoint union.)  I believe, however, that these two tensor products turn out to be the same, since any pair $(x,0)$ in $G(c)\times F(c)$ is equal to $(x,F(t)(0))$ where $t:0\to c$ is the unique map from 0 to $c$ in $C$, and hence gets identified in the tensor product with $(G(t)(x),0) = (0,0)$, which is the basepoint of $\bigvee_c G(c)\times F(c)$.  Thus when $C$ has a zero object, the ordinary tensor product over $C$ has the effect of automatically performing the smash product as well.
My question is: does this remain true for homotopy tensor products?  Suppose we consider $Set_\star$ as sitting inside the category $sSet_\star$ of pointed simplicial sets, and instead of the tensor product we take its homotopical replacement, which can be described as a two-sided bar construction.  We thus get two pointed simplicial sets $B^u(G,C,F)$ and $B^p(G,C,F)$ (for unpointed and pointed), defined by
$$B^u_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \times C(c_{n-1},c_n)\times\dots\times C(c_0,c_1)\times F(c_0) $$
and
$$B^p_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \wedge C(c_{n-1},c_n)\wedge\dots\wedge C(c_0,c_1)\wedge F(c_0) $$
There is a canonical quotient map $B^u(G,C,F) \to B^p(G,C,F)$, and the above observation (assuming it is correct) means that this map induces an isomorphism on $\pi_0$.  Is it a weak homotopy equivalence?
Edit: Reading over the answers, I realized that there are actually two different questions here.  Greg answered a question which isn't quite what I asked, but fortunately the question he answered is the one that I meant to ask, which is comparing $B^p(G,C,F)$ with the homotopy tensor product when considering $G$ and $F$ as functors landing in unpointed simplicial sets, so that the $\bigvee$s would actually become disjoint unions $\coprod$.

REPLY [6 votes]: I'll sketch an alternative approach. I've been on the lookout lately for ways to argue about homotopy (co)limits using universal properties instead of explicit constructions, so I like this question.
Let $Fun(C,Set_* )$ be the category of functors and let $Fun_* (C,Set_* )$ be the category of pointed functors. For a given choice of $G$, the weighted colimit functor $G\otimes -$ in the enriched sense is a composition $Fun_* (C,Set_* ) \rightarrow Fun(C,Set_* )\rightarrow Set_* $ , forgetful functor followed by the unenriched $G\otimes -$. This follows from the fact that its right adjoint is the composition of right adjoints $ Set_* \rightarrow  Fun(C, Set_* ) \rightarrow Fun_* (C,Set_* )$, $X\mapsto (c\mapsto Set_* (G(c),X))$ followed by $F\mapsto (c\mapsto (fiber(F(c)\rightarrow F(0))))$. 
Now make a derived-functor version of the same argument.