Fourier Transform - Dirac Delta

So the other answers are correct only for $x≠0$. In the sense of distributions, however, there is indeed a Dirac delta appearing in the result.

The first steps in your approach are correct. both functions $\frac{1}{q^2}$ and $\frac{(q·a)\,(q\cdot b)}{q^2}$ are indeed locally integrable functions and bounded outside a compact set, so are tempered distributions: one can take their Fourier transform in the sense of distributions (remark however that these functions are not in $L^1$ or in $L^2$, so one cannot take the Fourier transform in the usual sense).

In the sense of distributions (writing with abuse of notation the Fourier transform as an integral) $$ \frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q = -(a\cdot\nabla)(b\cdot\nabla)\, \frac{1}{4π|x|}. $$ Taking a first derivative of $1/|x|$, one obtains $-x/|x|^3$ (even in the sense of distributions) which is still a locally integrable function. However, taking two derivatives leads to a non locally integrable function, and so one has to be careful as you suspect. A formula that generalizes the formula of the Laplacian $-\Delta \frac{1}{4\pi|x|} = \delta_0$ is the Formula of the Hessian (where $\nabla^2 = \nabla\nabla$) $$ \nabla^2\left(\frac{1}{4\pi|x|}\right) = \frac{1}{4π} \,\mathrm{pv.}\,\frac{3x\otimes x - |x|^2 \,\mathrm{Id}}{|x|^5} - \frac{1}{3}\, \delta_0 \,\mathrm{Id} $$ where pv. denotes the principal value, see e.g. p.55 here. (In particular, if you sum the coordinates $(j,j)$ of this matrix, you obtain the formula for the Laplacian). Now remark that $(a\cdot\nabla)(b\cdot\nabla) = (a\otimes b):\nabla^2$ (if you prefer coordinates, $\sum_{ij} a_i\partial_i\, b_j\partial_j = \sum_{ij} a_ib_j\,\partial_i\partial_j$) and so $$ \begin{align} \frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q &= -(a\otimes b):\nabla^2\left(\frac{1}{4\pi|x|}\right). \\ &= \frac{1}{4π} \,\mathrm{pv.}\,\frac{|x|^2 (a·b) - 3\,(a\cdot x)(b\cdot x)}{|x|^5} + \frac{a\cdot b}{3}\, \delta_0 \end{align} $$

The physicist usually do not write the principal value, and so in shorter notations (with $r=|x|$) $$ \boxed{\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q = \frac{1}{4πr^3} \left(a·b - 3\,(a\cdot \tfrac{x}{r})(b\cdot \tfrac{x}{r})\right) + \frac{a\cdot b}{3}\, \delta_0} $$