Pythonista program to open a URL (given via iOS share sheet) in Safari. May require iOS 9. (Untested.)

QuickViewer.py

# coding: utf-8
# from: https://forum.omz-software.com/topic/2271/beta-suggestion-safariviewcontroller/9

from objc_util import *
import appex

SFSafariViewController = ObjCClass('SFSafariViewController')

def open_in_safari_vc(url):
    vc = SFSafariViewController.alloc().initWithURL_entersReaderIfAvailable_(nsurl(url), True)
    
    app = UIApplication.sharedApplication()
    
    if app.keyWindow():
        window = app.keyWindow()
    else:
            window = app.windows().firstObject()
    
    root_vc = window.rootViewController()
    
    while root_vc.presentedViewController():
        root_vc = root_vc.presentedViewController()
    root_vc.presentViewController_animated_completion_(vc, True, None)
    vc.release()

def main():
    if not appex.is_running_extension():
        print 'This script is intended to be run from the sharing extension.'
        return
    url = appex.get_url()
    if not url:
        print 'No input url'
        return
    open_in_safari_vc(url)
    
if __name__ == '__main__':
    main()