Yes. In fact you don't even need to use probability estimates for that, you can use the SVM's decision values directly (signed distance to the separating hyperplane).
When predicting a test instance $\mathbf{z}$ with an SVM, the following decision value is generated:
$$
\begin{align}
f(\mathbf{z}) &= \mathbf{w}^T\phi(\mathbf{z}) + b, \\
&= \sum_{i\in SV} \alpha_i y_i \kappa(\mathbf{x}_i,\mathbf{z}) + b.
\end{align}$$
This is a real value, e.g. $f(\cdot):\mathbb{R}^d \mapsto \mathbb{R}$. To obtain binary labels at the default threshold, the sign of this number is taken, e.g. $\hat{y}(\mathbf{z}) = \text{sign}(f(\mathbf{z}))$.
Most SVM software allows you to obtain $f(\mathbf{z})$ directly, without any extra work since this is done under the hood anyway.
Some software can additionally yield probability estimates, typically via Platt scaling. Obtaining probabilities is basically a matter of scaling $\mathbb{R}$ to $[0,1]$, for instance by running it through the logistic function with some scaling. Crucial here is that this is a monotonic transformation, e.g. this does not affect a ranking.