Intuitively, I would expect Q in this case to be a two dimensional vector space with basis 1,[a]. Is this correct? Projecting onto the orthogonal complement seems to give zero however.
The first part is correct. This may be easiest to see by considering the isomorphism, in the present case where $V$ is one dimensional, of $T(V)$ with $\mathbb{R}[x]$. Then $Q$ is $\mathbb{R}[x]/(x^2-x)$, which is 2-dimensional by the division algorithm.
The orthogonal complement of $I(S)$ in this case is not zero, but it is one dimensional and your concern is valid. Thinking again in $\mathbb{R}[x]$ for convenience, the condition that $\sum_k a_k x^k$ is orthogonal to $x^{m+2} - x^{m+1}$ for all $m\geq0$ says that $a_{m+2}=a_{m+1}$ for all $m\geq0$; thus, $a_k=0$ for $k\geq1$. Note that the conclusion is the same if we first complete to a Hilbert space by taking power series with square summable coefficients, so that unfortunately Andrew Stacey's desperate hope appears to be shattered. I have been assuming in this paragraph that $a$ (or $x$) has norm 1; it makes no difference for the conclusion in the non-completed case, but it simplifies the relation. (In other cases when $\dim(V)>1$ the Hilbert space quotient can be far too large for the vector space isomorphism to be possible, having dimension $2^{\aleph_0}$, so completion is probably not the way to go.)
This shows that there is no hope in general of using the standard inner product to define a norm on $T(V)/I(S)$. However, the problem in the above case arose because the ideal was not homogeneous; this forces relations on coefficients of different degree for elements of the orthogonal complement, thereby forcing too many coefficients to be zero. For this construction to work I recommend considering the case where $S$ is a set of homogeneous elements of $T(V)$ (i.e., each element of $S$ lies in one of the tensor powers $V^{\otimes k}$). I don't know what else to tell you in the nonhomogeneous case.
Gratuitous Add-on
The Hilbert space completion of the free algebra $T(V)$ is the full (or free) Fock space $F(V)$. The latter is not an algebra, but a Banach algebra completion of $T(V)$ can be obtained by first representing $T(V)$ on $F(V)$ by choosing an orthonormal basis $v_1,\ldots,v_n$ for $V$ and sending $v_j\in T(V)$ to the "creation operator" $S_j$ on $F(V)$ defined by $S_j(w)=v_j\otimes w$, and then closing the image in your favorite topology. These algebras were first studied by G. Popescu circa 1991 (but over $\mathbb{C}$ rather than $\mathbb{R}$ and with either norm or ultraweak closure), and a similar construction was earlier used by D. Evans to study the Cuntz algebra $\mathcal{O}_n$ in the 1980 paper "On $\mathcal{O}_n$".
