Quotient map not nullhomotopic

I have the following qual problem:

Let $M$ be a connected closed surface, not necessarily orientable, with an embedded closed disk $D$. Let $Q$ be the quotient space of $M$ by $\overline{M\setminus D}$. After showing $Q\cong S^2$ show that the quotient map $q: M \rightarrow S^2$ is not nullhomotopic.

If $M$ is orientable, I compute the degree of $q$ to be non-zero and deduce that the map is not nullhomotopic. However, if $M$ is non-orientable, then I cannot think of what to do. Since both the quotient map and a constant map descend to the quotient space $Q$, I would like to modify a nullhomotopy to descend to $Q$ and then arrive at a contradiction, but do not see how to pull that off. I have also thought about using, somehow, the oriented double cover of $M$.

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Do you know about modulo 2 degree? –  Micah Aug 26 '14 at 4:04
I think maybe you can use the fundamental polygon and the classification theorem to reduce it to $\mathbb{RP}^{2}$, then show this is impossible because if it is nullhomotopic then you get $\mathbb{RP}^{2}$ back. –  Bombyx mori Aug 26 '14 at 4:09

Consider the composite map $$(D.\partial D) \rightarrow (M,M-D) \rightarrow (S^2, x_0)$$ It suffices to show that $(D.\partial D) \rightarrow (S^2, x_0)$ is not nullhomotopic. For example it induces an isomorphism in homology $H_2(D.\partial D) \rightarrow H_2(S^2, x_0)$.
Maybe the following helps to get the conceptual idea: since you collapse everything in $M-D$ to a point, you are not further interested in it. Hence you can use relative homology to ignore that part. If you have nice objects like manifolds you can usually use the homology of the quotient space to compute relative homology and also you can use excision. Think about it! –  Dan Aug 26 '14 at 7:27