ListingController.cs

public class ListingPageController : RenderMvcController
{
    public override ActionResult Index(RenderModel model)
    {
        // Do stuff here

        return CurrentTemplate(new ListingViewModel(model)
        {
            CurrentListing = listing
        });
    }
  }

ListingPage.cshtml

@using MyNamespace.Web.ViewModels
@inherits Umbraco.Web.Mvc.UmbracoViewPage<ListingViewModel>
@{
    Layout = "Master.cshtml";
}

<h1>@Model.CurrentListing.Name</h1>

ListingViewModel.cs

// A issue with Umbraco hijacked routes and custom view models
// means our viewmodel currently has to inherit from RenderModel
public class ListingViewModel : RenderModel
{
    public ListingViewModel(RenderModel model)
        : this(model.Content, model.CurrentCulture)
    { }

    public ListingViewModel(IPublishedContent content, CultureInfo culture) 
        : base(content, culture)
    { }

    public ListingViewModel(IPublishedContent content) 
        : base(content)
    { }

    public Listing CurrentListing { get; set; }
}

If you return a view like you do in normal Mvc, then it's perfectly fine to return a custom model that doesn't inherit from RenderModel. But then you have to specify the view yourself, which I'm guessing you would like to avoid?

public class ListingViewModel : RenderMvcController
{
    public override ActionResult Index(RenderModel model)
    {
        ...

        return View("SomeListingView", new ListingViewModel());
    }    
}

------------------

@inherits UmbracoViewPage<ListingViewModel>
@{
    Layout = "~/Views/Shared/_Layout.cshtml";
}
<h1>Listing view</h1>

At least, I have that working perfectly.