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My thought is no. The formula $a=\frac{v^2}{r}$ only applies to an object in uniform circular motion. If that motion is decreasing, I think it may still have constant VELOCITY towards the center of the circle though. But my math isn't good enough to prove or disprove that thought. I tried using a similar triangle argument with vectors, but I got nowhere with it. Anyone up for the challenge? I wasn't sure if this is more of a physics question or a math question since it concerns the motion of objects and the mathematical derivation of that.

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    Yes, of course. If it doesn't accelerate towards the center, it moves in a straight line. – 2017-01-18

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The acceleration of a circular movement decomposes into two orthogonal components. One component points toward the center and the other is tangential to the circle. So the acceleration points toward the center if and only if the tangential component is zero. This component is proportional to the angular acceleration (i.e. the change in angular velocity). This component will therefore not be zero if the object is decelerating along the circle. In a formula: $$\frac{\partial^2}{\partial t^2}(\cos \varphi, \sin \varphi) = -\dot{\varphi}^2\cdot(\cos \varphi, \sin \varphi) + \ddot{\varphi}\cdot(-\sin \varphi, \cos \varphi).$$ Here $\varphi=\varphi(t)$ is the angle as a function of time.

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    Good argument, $+1$. I was going to give an explicit example, much too complicated. Here’s a physical argument: if the acceleration is towards the center, the angular momentum is constant. Certainly a slowing-down object moving in a circle has decreasing angular momentum. – 2017-01-18
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Define the magnitude of the velocity as

$|v(t)|=f(t)$

According to the formula $w=rf(t)$, where $w$ is angular velocity, $f(t)$ is linear velocity and $r$ is the radius of the circle, we can have the relation

$w=\frac{d\theta}{dt}= rf(t)$

Assume that the circle is centred at the origin and we start moving from the most top point of the circle, counter-clockwise. The velocity vector at time $t$ is

$v(t)=\begin{pmatrix}-f(t)cos\theta \\ -f(t)sin\theta\end{pmatrix}$

$\theta$ is a function of $t$.

To find acceleration you need to calculate $a(t)=\frac{dv}{dt}$. As $v(t)$ is a vector, we can differentiate with respect to each component separately.

$a(t)=\begin{pmatrix}\frac{d(-f(t)cos\theta)}{dt} \\ \frac{d(-f(t)sin\theta)}{dt}\end{pmatrix}=\begin{pmatrix}\frac{-d(f(t))}{dt} cos\theta + rf(t)sin\theta \\ \frac{-d(f(t))}{dt} sin\theta -rf(t)cos\theta\end{pmatrix}$

This vector should be in the same direction as $\begin{pmatrix}-sin\theta\\ cos\theta\end{pmatrix}$, if we want the acceleration to be towards the centre. For a motion, with uniform speed, $\frac{d(f(t))}{dt}=0$ and $f(t)$ would be constant and, therefore, $a(t)=-r\begin{pmatrix}-sin\theta\\ cos\theta\end{pmatrix}$. However, if $\frac{d(f(t))}{dt}\neq 0$, in general, we do not have the acceleration directed towards the centre, all the time.