Restating my comment (after looking at the formulas a second time), it should satisfy the Leibniz rule, which someone can reword as being a graded derivation. Sketch of proof:
Cup product works on mixed relative cohomologies, $H^p(X,A_1)\times H^q(X,A_2)\to H^{p+q}(X,A_1\cup A_2)$, so we can take $A_1=A_2=A$ or $A_1=A_2=\varnothing$ or $A_1=A$ and $A_2=\varnothing$ and vice versa. The coboundary map $H^n(A)\to H^{n+1}(X)$ is obtained by taking co-chains on $A$ and viewing them as co-chains on $X$ which vanish on $X-A$ and then pre-composing with the differential $C_{n+1}(X)\to C_n(X)$, i.e. it is obtained directly from the co-differential. If you work out the formula for the co-differential of the cup product of co-chains (which is the Leibniz rule), this should respect the values of the relative co-chains.
So in your notation, $\partial_{p+q}(\alpha\smile\beta) = \partial_p\alpha\smile\beta\pm \alpha\smile\partial_q\beta$, and I'll leave you to the signs (that's a phrase I hear way too often in math).
Addendum: There is a "stability property" written in Dold's Lectures on Algebraic Topology (and other books) which states a special case, $\partial\alpha\smile\eta=\partial(\alpha\smile i^\ast\eta)$, where $\alpha\in H^p(A)$ and $\eta\in H^q(X)$ and $i:A\to X$ is inclusion. This is in agreement with what I wrote, because $\delta i^\ast=0$ by exactness.