How to change CSS when it's ng-disabled?

What Toress answered should work fine but you don't need the help of AngularJS here at all (a native implementation & usage is always best).

You can make use of CSS3 since you already have a class on it. Example:

input.save-changes {
    /* some style when the element is active */
}

input.save-changes[disabled] {
    /* styles when the element is disabled */
    background-color: #ddd;
}

Edit: You can immediately test it on this page of StackOverflow. Just inspect the blue button element and put the disabled attribute and see it's CSS.

.save-changes {
  background-color: red;
  padding: 7px 13px;
  color: white;
  border: 1px solid red;
  font-weight: bold;
}
.save-changes[disabled] {
  background-color: #FF85A1
}

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>

<div ng-app ng-init="PayoutEnabled = true">
  <a href="#" ng-click="PayoutEnabled = !PayoutEnabled">
  {{PayoutEnabled ? 'Disable' : 'Enable'}} the below button</a>
  <br>
  <br>

  <input class="save-changes" type="submit" value="PAYOUT" ng-disabled="PayoutEnabled == false" />
</div>

use ng-class

<input type="submit" value="@Translator.Translate("PAYOUT")" class="btn-block 
secondary-button save-changes padding-8" ng-disabled="PayoutEnabled==false" 
ng-click="PayOut()" ng-class="{'diabled-class': !PayoutEnabled}" />

this will add css class diabled-class to the input when PayoutEnabled is false (!PayoutEnabled is true).

AngularJS adds pseudo-class disabled when ng-disabled is false so i think here is the simplest solution to refer to disabled button :

button:disabled {
    color:#717782;
}