This answers the first (simple) half of the question, asking just about a smooth map. In fact, you've already given an answer to it, in some sense. Apply the map $f: re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$ to a unit disk that doesn't contain $(0,0)$, say radius $1$ disk $D$, centred at $(2,0)$. Then, the image $f(D)$ is contained in the disk of radius $3\sigma_1$, centred at $(0,0)$. So if $\sigma_1<\frac{1}{3}$, the map $f+(2,0)$ sends $D$ to $D$ and has the desired properties.
PS. Concerning the second part of the question about a diffeomorphism, I can't give an example, but can give something that looks almost like an example (though one can show that this can not be perturbed to an example by a perturbation which is $C^1$-small close to the boundary of the disk). I'll construct a one parameter family of maps $f_c:D\to D$ (two maps $f_c$ for each $\sigma_1\in (0,1)$). These maps are diffeos on the complement to $(0,0)\in D$, and have all required properties on $D\setminus (0,0)$, though they are not differentiable at $(0,0)$. In the radial coordinates the formula is as follows:
$$f_c: (r,\theta)\to r(\cos(\theta+c\log(r)), \sin(\theta+c\log (r))),\;\; c\in \mathbb R.$$
In order to see that these maps satisfy the necessary conditions, it is enough to notice that they have the following two properties:
1) Each circle $x^2+y^2=r^2$ is sent isometrically to itself.
2) Each radial segment $(\cos(\theta)t, \sin(\theta)t)$ ($t\in [0,1]$) is sent to a spiral $e^s(\cos(\psi+cs),\sin(\psi+cs))$ parameterized by $s$ (where $\psi$ is a constant that depends on $\theta$).
All the conditions are satisfied because the family of spirals and the family of circles form the same angle at all points in $D\setminus (0,0)$, and the map is obviously a symplectomorphism and has unit norm on all the circles. For each $\sigma_1<1$ correspond exactly two $c$ (that differ by a sign).
PPS. 22/02/2020 I would like to propose one more statement (of which I am very happy), concerning the maps with $\sigma_1\in (\frac{1}{2},1]$.
Lemma. Suppose that we have a smooth map $f: D\to \mathbb R^2$ from the unit disk to $\mathbb R^2$ with fixed $\sigma_1<1$ and $\sigma_2=\frac{1}{\sigma_1}$. Then $f$ is univalent (i.e. a diffeo on its image) in case $\sigma_1>\frac{1}{\sqrt{2}}$.
The main tool of the proof is the isomperimetric inequality that says that a simple closed curve $\eta$ on $\mathbb R^2$ bounds area nor more than $\frac{l(\eta)^2}{4\pi}$. It is also surprising, that the constant $\frac{1}{\sqrt{2}}$ is exact, i.e. for $\sigma_1<\frac{1}{\sqrt{2}}$ the map doesn't need to be univalent!
Proof. Assume the converse. Denote by $D_r\subset D$ a disk of radius $r\le 1$ centred at $(0,0)$. Clearly, for $r$ small enough the restriction map $f: D_r\to D$ is a diffeomorphism onto its image. Hence there is a minimal $t\in (0,1]$, such that this map is not a diffeo on its image. Let $S_t$ be the boundary of $D_t$ (a circle of radius $t$). Clearly, the curve $f(S_t)$ touches itself at some point. There could be more than one point where it touches itself, but the argument will not change, so we will assume that $f(S_t)$ self-touches once.
Let $x$ and $y$ be the two points in $S_t$ such that $f(x)=f(y)$. Let $(xy)$ and $(yx)$ be the two arcs into which $x$ and $y$ cut $S_t$. Without loss of generality we assume that the arc $(xy)$ is longer than $(yx)$. Denote by $\gamma_{xy}$ and $\gamma_{yx}$ the images $f((x,y))$ and $f((y,x))$. Both these images are simple closed loops. Let's prove first that the loop $\gamma_{xy}$ contains the loop $\gamma_{yx}$ in its interior.
Indeed, assume the converse. Note that by definition, the length $l(\gamma_{yx})$ satisfies
$$l(\gamma_{yx})\le \frac{1}{\sigma_1}l([yx])\le \sqrt{2}\cdot \pi t.$$
One the other hand, the curve $\gamma_{yx}$ encloses the whole image $f(D_t)$ of $D_t$ and the disk bounded by $\gamma_{xy}$, that doesn't belong to $f(D_t)$. So, since $f$ is area preserving, we see that $\gamma_{yx}$ encloses area more than $\pi t^2$. This contradicts the isomperimetric inequality.
We conclude, that $\gamma_{xy}$ encloses $\gamma_{yx}$, and moreover $l((xy))>\pi t$.
Let's go on to get a contradiction. Let $z$ be the midpoint of the chord $[xy]$ in $D_t$ that joins $x$ and $y$ (don't confuse it with the curvy arc $(x,y)$ that lies in $S_t$!). Let $z$ be the midpoint of $[xy]$. Consider the circle $S_z$ centred at $z$ that passes through $x$ and $y$. Then one half of this circle lies inside $D_t$. Denote this half-circle by $\eta$. Obviously, $l(\eta)=\pi\frac{l([xy])}{2}$. Denote further by $D_t'$ the connected component of $D_t\setminus \eta$ that contains the shorter arc $(y,x)$ of $S_t$. Note finally, that
$${\rm area}(D')>\frac{1}{2}\pi\left (\frac{l([xy])}{2}\right)^2,$$
since $D'$ contains a half-disk of radius $\frac{l([x,y])}{2}$ centred at $z$.
Now, to get the contradiction, we apply again the isoperimetric ineqaulity, this time to the disk bounded by the simple curve $f(\eta)$. By construction, the disk bounded by $f(\eta)$ contains in its interior the image $f(D')$, so it has area more than ${\rm area}(D')$. At the same time the length of $f(\eta)$ is less than $\frac{\pi}{\sqrt{2}} l(xy)$. END of proof.
Moral. If there are self-maps with $\sigma_1>\frac{1}{\sqrt{2}}$, they are diffeos... But I still don't know if such diffeos exist:)
PPPS. 25/02. I would like to address Asaf's question, stated in the comments. Namely, is it possible to find any symplectomorphism $f:D\to D$ ($D$ is the unit disk) that has distinct singular values at any points. It seems to me that the infinitesimal version of this question is equivalent to the following funny question:
Question. Is it possible to construct a smooth function $H$ of $D$, vanishing at the boundary of $D$, and such that the system of equations
$$\frac{\partial^2 H}{\partial x^2}-\frac{\partial^2 H}{\partial y^2}=0=\frac{\partial^2 H}{\partial x\partial y}$$
has no solutions in the disk?
Indeed, if we consider the Hamiltonian flow for $H$, satisfying the above conditions, I believe that for small time the flow map will have distinct singular values. What is good about this system, is that it should not be hard to program (for someone who unlike me knows how to do this), to look for examples. Namely, one can fix a degree $d>0$ and consider all polynomials $H=(x^2+y^2-1)P_d$, where $P_d$ is a poly of degree at most $d$. The space of such polynomials (for all $d$) will probably be dense in the space of all smooth functions. So if there is a counterexample to the question, it should be possible to find it among polynomials. Maybe its degree will not be very large (according to Arnold's "Topological Economy Principle in Algebraic Geometry", ). Or, on the contrary, if there is no counterexample among polynomials, there will be no among all functions, this would be quite exiting. It is easy to check that for $d=1$ indeed there is no counterexamples.