Here is a spelled out version of Tom Goodwillie's comment.

Let $\mathcal{M}$ be the category of connected smooth manifolds and smooth maps, let $\mathcal{M}_1$ be the subcategory with the same objects whose morphisms are the diffeomorphisms, and let $J\colon\mathcal{M}_1\to\mathcal{M}$ be the inclusion. Suppose we have a functor $U\colon\mathcal{M}_1\to\mathcal{M}$ and a natural map $p\colon UM\to JM$ that is a universal cover for all $M$. Consider $S^1$ as the usual subspace of $\mathbb{C}$, and choose a point $a\in p^{-1}\{1\}\subset U(S^1)$. For $z\in S^1$ we can define $\mu_z\in\mathcal{M}_1(S^1,S^1)$ by $\mu_z(u)=zu$, and then define $s(z)=U(\mu_z)(a)\in U(S^1)$. This defines a section $s$ of the map $p\colon U(S^1)\to S^1$. If we make enough additional assumptions to ensure that $s$ is continuous, then we arrive at a contradiction.

I think that in fact no additional assumptions are needed, but that needs a slightly different approach. We can identify $S^1$ with $\mathbb{R}P^1$, and then we have an action of the group $G=PSL_2(\mathbb{R})$. Let $H$ be the upper triangular subgroup, which is the stabiliser of the basepoint $1\in S^1$. For $h\in H$ there is a unique $h'\colon U(S^1)\to U(S^1)$ with $ph'=hp$ and $h'(a)=a$. The map $Fh$ need not obviously fix $a$ so it need not coincide with $h'$, but it must have $Fh=\phi(h)\circ h'$ for some deck transformation $\phi(h)$. The group of deck transformations can be identified with $\pi_1(S^1,1)=\mathbb{Z}$, and $H$ acts on this in a natural way (independent of the supposed existence of $U$). Using the connectivity of $H$ we see that this action is trivial. I think it follows that $\phi\colon H\to\mathbb{Z}$ is a homomorphism, but any element $h\in H$ has $n$'th roots for all $n>0$, and this forces $\phi$ to be trivial, so $Fh=h'$ for all $h$. This proves that $Fh$ depends continuously on $h$ for $h\in H$. Moreover, one can find $h_z,k_z\in H$ such that the entries are rational functions of $z$ and $\mu_z=h_z\mu_{-1}k_z$. It follows that $F(\mu_z)$ depends continuously on $z$ except possibly at finitely many values of $z$. These possible exceptions can then be removed by an auxiliary argument with the group structure.

[UPDATE: As Tom Goodwillie points out, this is much more complicated than necessary and misunderstands the line of argument that he had in mind. Still, it has some interesting features so I will leave it here.]

Let $\mathcal{M}$ be the category of connected smooth manifolds and smooth maps, let $\mathcal{M}_1$ be the subcategory with the same objects whose morphisms are the diffeomorphisms, and let $J\colon\mathcal{M}_1\to\mathcal{M}$ be the inclusion. Suppose we have a functor $U\colon\mathcal{M}_1\to\mathcal{M}$ and a natural map $p\colon UM\to JM$ that is a universal cover for all $M$. Consider $S^1$ as the usual subspace of $\mathbb{C}$, and choose a point $a\in p^{-1}\{1\}\subset U(S^1)$. For $z\in S^1$ we can define $\mu_z\in\mathcal{M}_1(S^1,S^1)$ by $\mu_z(u)=zu$, and then define $s(z)=U(\mu_z)(a)\in U(S^1)$. This defines a section $s$ of the map $p\colon U(S^1)\to S^1$. If we make enough additional assumptions to ensure that $s$ is continuous, then we arrive at a contradiction.

I think that in fact no additional assumptions are needed, but that needs a slightly different approach. We can identify $S^1$ with $\mathbb{R}P^1$, and then we have an action of the group $G=PSL_2(\mathbb{R})$. Let $H$ be the upper triangular subgroup, which is the stabiliser of the basepoint $1\in S^1$. For $h\in H$ there is a unique $h'\colon U(S^1)\to U(S^1)$ with $ph'=hp$ and $h'(a)=a$. The map $Fh$ need not obviously fix $a$ so it need not coincide with $h'$, but it must have $Fh=\phi(h)\circ h'$ for some deck transformation $\phi(h)$. The group of deck transformations can be identified with $\pi_1(S^1,1)=\mathbb{Z}$, and $H$ acts on this in a natural way (independent of the supposed existence of $U$). Using the connectivity of $H$ we see that this action is trivial. I think it follows that $\phi\colon H\to\mathbb{Z}$ is a homomorphism, but any element $h\in H$ has $n$'th roots for all $n>0$, and this forces $\phi$ to be trivial, so $Fh=h'$ for all $h$. This proves that $Fh$ depends continuously on $h$ for $h\in H$. Moreover, one can find $h_z,k_z\in H$ such that the entries are rational functions of $z$ and $\mu_z=h_z\mu_{-1}k_z$. It follows that $F(\mu_z)$ depends continuously on $z$ except possibly at finitely many values of $z$. These possible exceptions can then be removed by an auxiliary argument with the group structure.