**UPDATE**: the following argument is wrong, see the comments.

If $\mathcal{C}l$ admits a right adjoint then it preserves colimits, and coproducts in particular. Now, in your category of quadratic vector spaces, the coproduct of $(V, Q)$ and $(V', Q')$ is $(V \oplus V', Q \oplus Q')$; for associative algebras $A$ and $A'$, its coproduct is given by tensor product over $\mathbb{R}$. Hence, it is necessary that $$\mathcal{C}l(V \oplus V', Q \oplus Q') \cong \mathcal{C}l(V, Q) \otimes_{\mathbb{R}} \mathcal{C}l(V', Q')$$

Here's a counterexample: take $V = V' = \mathbb{R}$ with $Q = Q' = -1$. By the [classification of Clifford algebras][1], we know that $\mathcal{C}l(\mathbb{R}, -1) \cong \mathbb{C}$ and $\mathcal{C}l(\mathbb{R}^2, \mathrm{diag}(-1,-1)) \cong \mathbb{H}$. It is now enough to observe that $$\mathbb{H} \not\cong \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} $$


  [1]: http://en.wikipedia.org/wiki/Classification_of_Clifford_algebras