Let $C$ be a fusion category with simple objects $X_i$. If $X$ is an object, then $X \otimes X_i = \bigoplus N_{ij} X_j$ for some matrix $N_{ij}$ of non-negative integers. The largest non-negative real eigenvalue of this matrix is called the Frobenius-Perron dimension $\dim X$ of $X$. It is preserved by exact tensor functors (Corollary 4.6 in the linked paper) and coincides with the ordinary dimension for finite-dimensional vector spaces. It follows that a fusion category with an object whose Frobenius-Perron dimension is not an integer does not admit an exact tensor functor to $\text{FinVect}$. If I'm not mistaken, tensor functors preserve dualizable objects, so any tensor functor to $\text{Vect}$ must land in $\text{FinVect}$ anyway, and the conclusion follows.

In particular there is a fusion category with an object $X$ satisfying $X^{\otimes 2} \cong 1 \oplus X$ whose Frobenius-Perron dimension is $\frac{1 + \sqrt{5}}{2}$.

**Edit:** Here is maybe a simpler way of putting it. Associated to any fusion category $C$ is its Grothendieck ring, and an exact tensor functor between fusion categories gives a homomorphism of Grothendieck rings. So it suffices to find a fusion category whose Grothendieck ring does not admit a homomorphism to $\mathbb{Z}$, and clearly a fusion category with an object $X$ satisfying $X^{\otimes 2} \cong 1 \oplus X$ has this property.

**Edit #2:** It seems that in the literature a Tannakian category is required to be *symmetric* monoidal. In this case, ignore everything I said above.

Deligne (see e.g. this exposition by Ostrik) showed that over an algebraically closed field of characteristic $0$ together with some other mild hypotheses, which hold in particular for symmetric fusion categories, any such category admits a *super* fiber functor to $\text{sVect}$. From here one can classify symmetric fusion categories in terms of finite groups; see section 2.12 of On braided fusion categories I by Drinfeld, Gelaki, Nikshych, and Ostrik. $\text{sVect}$ itself does not admit an ordinary fiber functor, as Angelo pointed out; the obvious functor doesn't preserve symmetries.