No, and a simple example goes as follows:
$$
\begin{split}
K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) 
& =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\
& \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N
\end{split}
$$
where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in [this answer](https://mathoverflow.net/questions/349057/question-about-functional-derivatives/349584#349584) is not.

The standard kernel theorem ([1] chapter 1, §1.3 pp. 11-20 and §3.5 pp. 73-79) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$-linear functionals on particular function spaces.

**References**

Gel’fand, I. M.; Vilenkin, N. Ya., _Generalized functions. Vol. 4: Applications of harmonic analysis_, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), [MR0173945](http://www.ams.org/mathscinet-getitem?mr=MR0173945), [Zbl 0136.11201](https://zbmath.org/?q=an%3A0136.11201).